Why is torque a cross product? If I'm not wrong, torque is perpendicular to both the radius and force i. e. It is along the axis of rotation. Questions that arise are- why do we consider the length between the axis/point of rotation while calculating torque? More importantly why is torque a cross product? 
 A: (a) "why do we consider the length between the axis/point of rotation while calculating torque?"
We can calculate torque about any point, O, that we choose; it doesn't have to be a physical axis of rotation. But it's often more useful to calculate torque about a possible physical rotation axis, for example when thinking about what torque we need to apply on a nut with a spanner (wrench) in order to undo it. As for why length (or, specifically, perpendicular distance from O to the line of action of the force) comes into the definition, just think about trying to undo that nut!
(b) "why is torque a cross product?" 
In vector notation we define the torque about O due to a force $\vec{F}$ acting at a point displaced by $\vec{r}$ from O to be $\vec{r} \times \vec{F}$. The magnitude, $|\vec{r} \times \vec{F}|$of this torque vector is exactly equivalent to the "force $\times$ perpendicular distance" definition that I quoted from in (a). The direction of $\vec{r} \times \vec{F}$ is at right angles to the plane containing  $\vec r$ and $\vec F$ and therefore tells you the alignment of the (possibly imaginary) axis about which the torque, acting on a nut, would turn it! [In fact. with the usual 'right handed' convention for defining cross product, if you point the thumb of your right hand in the direction of $\vec{r} \times \vec{F}$, the fingers of that hand will tend to curl round in the sense that the nut will turn!]
A: Torque is defined as $\quad\vec{\tau}=\frac{d\vec{J}}{dt}$ where $\vec{J}$ is the angular momentum of the object.  The angular momentum is defined as $\vec{J}=\vec{r}\times \vec{P}$.  Then 
$$
\vec{\tau}=\frac{d\vec{J}}{dt}=\frac{d(\vec{r}\times \vec{P})}{dt}=\frac{d\vec{r}}{dt}\times\vec{P}+\vec{r}\times\frac{d\vec{P}}{dt}
$$
but
$$
\frac{d\vec{r}}{dt}\times\vec{P}=\frac{d\vec{r}}{dt}\times m\vec{v}=\frac{d\vec{r}}{dt}\times m\frac{d\vec{r}}{dt}=0
$$
so
$$
\vec{\tau}=\vec{r}\times\frac{d\vec{P}}{dt}=\vec{r}\times\vec{F}
$$
which is the answer to the question.
A: It doesn't have to be thought of as cross product.  It's just very convenient to think of it that way, so we teach it first.  Indeed, even when I apply it in my job, I think of it as a cross product.
But first, your question about why the lever arm appears in the equations.  Informally, we need to account for the length because a longer lever arm gives you more mechanical advantage.  You can test this, yourself, with a wrench.  Try to tighten a bolt holding the wrench right up near the head, then hold the wrench further out near the end, giving yourself a longer lever arm, and try to tighten it.  You'll find you can tighten the bolt much better if you have a longer lever arm.
As for a mathematical explanation,  you can show it using conservation of momentum and angular momentum. Construct any scenario using forces and show that momentum is conserved (it should be!).  Now, pick any point as the "center" of your rotation, and calculate torques.  You'll find that angular momentum is conserved.  If you defined torque without the radius term, you'd find angular momentum would not be conserved.  In fact, it turns out that if you have forces and conservation of momentum, you can always derive torques and conservation of angular momentum.  And if you have torques and conservation of angular momentum, you can always derive the forces and the conservation of momentum!  They're sort of duals of one another.
If you want to go further than that, many years from now you'll learn Lagrangian Mechanics and Nother's Theorem.  You'll learn that the conservation of momentum is a very fundamental concept directly tied to the fact that our laws of physics are the same in all directions.  Rotate an experiment, and the laws of physics will stay the same.  There is no privileged direction where the laws of physics are "correct."
As for why torque is perpendicular to the force and the lever arm, that is actually just an artifact of mathematics, nothing more.  When you get deeper into Lagrangian Mechanics, what you'll find is that this angular momentum is just one specialized case of a wider concept called "generalized angular momentum."  In generalized angular momentum, the equivalent of torque is formed by the exterior product, r ∧ F.  This is known as a bivector, as opposed to a normal vector.  This works in any number of dimensions.
The exact definition of these bivectors is a bit of a pest to work with:

The exterior algebra Λ(V) of a vector space V over a field K is defined as the quotient algebra of the tensor algebra T(V) by the two-sided ideal I generated by all elements of the form x ⊗ x for x ∈ V (i.e. all tensors that can be expressed as the tensor product of a vector in V by itself).

What a whopper!  However, we're really lucky that we live in 3 dimensions.  As it turns out, when you crank out one of these bivectors in 3 dimensions, and look at how it behaves, a curious convenience shows up.  They behave exactly the same as cross products.  A bivector is not a vector, but it turns out these 3 dimensional bivectors have the same mathematical properties as cross products (which are a 3 dimensional concept).
Incidentally, this is also why we have to choose the right hand rule convention.  Bivectors can be calculated without such a convention, but when you map them into vectors using the cross product, there's two choices you can make -- left handed or right handed.  As long as you always choose one, the result is consistent.
Thus, for reasons that should be obvious, we choose to teach torque as a vector defined by r x F, rather than a bivector, r ∧ F.  It's a whole lot simpler!  But it comes with a price.  The vector r x F has a "direction," since it's a vector.  That direction is perpendicular to the force and the lever arm.  The bivector didn't have this particular concept of direction.  The concept of bivector direction is more nuanced, and more intuitively related to the direction of the force and the direction of the lever arm.
And so, you have your reason for the torque being "perpendicular."  It really doesn't have anything to do with physics, as much as it has to do with avoiding having to teach you advanced vector algebra to do basic physics.  You get the right answer using the cross product, because cross products and 3 dimensional exterior products operate the same.
A: Perhaps the best way to explain torque is with respect to the use of a common tool: The wrench.
If we want to tighten or loosen a bolt we may use a wrench. The first thing we realize is if the bolt is frozen in place, we are more likely to free it if we use a wrench with a longer arm. This is because the force we apply to the wrench arm produces more torque, or turning force, the longer the wrench arm. We commonly call this getting more "leverage".
First, you may notice if you try and turn the wrench with your force at an angle with the wrench arm, it is harder to turn the bolt than when it is perpendicular to the arm. In short, your torque is maximized when the angle of your applied force is 90 deg with the arm. In the other extreme, if you pulled or pushed axially on the arm, i.e., at an angle of zero, the bolt won't turn at all. For any angle in between it is only the component of the force perpendicular to the arm of the wrench that produces torque. Given an angle θ between the force $F$ and the arm of the wrench the component of force perpendicular to the arm has a magnitude of $F$ sin θ. 
Next consider that torque is a vector quantity and that a vector quantity has direction. The torque is in the direction of the angular velocity that could be produced if there is a net torque about an axis, such as that causing a wheel to rotate.  Since the only fixed unique direction of the rotating wheel is its axis of rotation, that axis is a logical choice for the general orientation of the torque and angular velocity vector, leaving two choices about the direction of the vector.  It is then customary to use the right hand rule to specify the direction.  This direction can also be visualized in terms of the direction that our bolt with a right hand thread would move when turned.  Looking down on the plane containing the wrench and the force applied to it, turning the bolt counter-clockwise causes the bolt to move up.  We call this the direction of positive torque.
Putting it together, the torque vector is the cross product of the force $F$ times the moment arm d (length of the wrench arm from the center of rotation to the point of application of force) or 
$$\vec{T}=\vec{F} \times \vec{d}$$
and 
$$|\vec{T}|= |\vec{F}| |\vec{d}| \sin \theta.$$
Where the direction of $\vec{T}$ is perpendicular to the plane containing $\vec{d}$ and $\vec{F}$. By convention, using the right hand rule, where the fingers curl in the direction of turning the thumb points in the direction of positive torque. 
Hope this helps.
A: This is a nice question. I agree that the standard textbook reflection on this problem generally leaves you wanting more (but this is perhaps unavoidable).
The basic question one grapples with at the start is $-$ how come suddenly we've to switch to torques and angular momentum when it comes to rigid bodies, as opposed to mere forces in case of point particles? A beautiful answer to this can be found in Synge & Schild, Tensor Calculus.
The real reason for this lies in the kinematics of rotation (i.e. what are the degrees of freedom and how does one go about describing them).
The basic logical idea here is that, in $D$ dimensions, a rigid body has $\frac{D(D+1)}{2}$ degrees of freedom $-$ $\frac{D(D-1)}{2}$ rotational & $D$ translational. If you don't worry about translational degrees of freedom for now, you can argue using basic linear algebra that rigid Euclidean motion can be described using orthogonal matrices (i.e. if you sit in an inertial frame and use cartesian coordinates). In odd spatial dimensions, this can be thought of as rotation of the body around some axis (for example in 3D) passing through the stationary point (cf. eg. Goldstein).
Furthermore, it then turns out that infinitesimal rotations can be described using real antisymmetric matrices (in a fancier parlance, one says that the Lie Algebra of the orthogonal matrices is spanned by real antisymmetric matrices). The components of this matrix describe the (cartesian) tensor of angular velocity $-$ note the number of independent components is $\frac{D(D-1)}{2}$ in number, reflecting $\frac{D(D-1)}{2}$ rotational degrees of freedom. If you know the dynamics of these components, you will be able to predict the orientation of the rigid body.
What you're then really looking for are $\frac{D(D-1)}{2}$ rotational equations of motion to describe the cartesian tensor of angular velocity. You can do it by using Newton's laws or the D'Alembert's principle (subject to the fact that the rigid body can only move in specific ways as allowed by the degrees of freedom). In either case, you will arrive at the equation
$$ \frac{d}{dt}L_{ij} = \tau_{ij}, $$
where $L_{ij} = \sum m(x_i p_j - x_j p_i)$ and $\tau_{ij} = x_i F_j - x_j F_i$.
$L_{ij}$ here is called the cartesian tensor of angular momentum, and $\tau_{ij}$ is called torque (also a cartesian tensor). In Newtonian mechanics, this includes an extra assumption that the internal forces in the rigid body between two particles is along the line joining them and hence doesn't contribute to the torque on the rigid body. One then often writes the equations of motion as,
$$ \frac{d}{dt}L_{ij} = \tau_{ij}^{\text{ext}}, $$
signifying that only external torques matter.
As already described by @CortAmmon, this is basically equivalent to the more elegant language of differential forms (that are naturally equipped with an antisymmetric structure).
In 3 dimensions it so happens that the above equation, which is a second order cartesian tensor equation of antisymmetric tensors, can be expressed as a vector equation in terms of its Hodge dual (assuming Euclidean metric on space). The result is that one talks of the pseudo-vector of angular momentum and the pseudo-vector of torque.

There is also a fancier version of D'Alembert principle, which goes by the name Lagrangian mechanics, wherein one can use the deep insights of Emmy Noether between symmetries & conservation laws. As @BenCrowell has pointed out in one of the comments that this would predict that torque free motion of a rigid body conserves angular momentum (this actually entails the assumption that the equal & opposite internal forces are acting along the line joining the internal parts so that internal potential energy is independent of the orientation of the rigid body). With this one can think of angular momentum as the conserved quantity in case of free motion of rigid bodies, whose derivative in the general case is equivalent to the torque equation (with torques expressed in terms of forces & the points where they are applied).
In either case, as claimed earlier, the real answer lies in the kinematics of rotation $-$ namely, the fact that infinitesimal rotations are described using real antisymmetric matrices (in cartesian coordinates).

A High School Explanation
Let's pick up the energy argument from Feynman lectures.
Consider the rotation of a body with one point fixed in 3D. One can define the velocity field as $\mathbf{v} = \mathbf{\omega}\times\mathbf{r} $, where $\mathbf{\omega}$ is the angular velocity (pseudo-vector) and the origin of the coordinate system is centered at the fixed point.
The kinetic energy of the body is then given by,
$$ K = \sum \frac{1}{2} m (\mathbf{\omega}\times \mathbf{r})^2. $$
And the rate of change of kinetic energy is simply,
$$\frac{d}{dt} K = \sum \Bigg[m \frac{d}{dt} (\mathbf{\omega}\times \mathbf{r})\Bigg] \cdot (\mathbf{\omega}\times \mathbf{r}) = \sum \Bigg[m \frac{d}{dt} \big[\mathbf{r}\times(\mathbf{\omega}\times \mathbf{r})\big]\Bigg] \cdot {\omega} = \frac{d}{dt} \mathbf{L} \cdot \mathbf{\omega} $$
The power generated by external forces would be equal to $\frac{d}{dt} K$,
$$ P = \sum \mathbf{F}_i.(\mathbf{\omega}\times\mathbf{r}_i) = \sum \mathbf{\omega} \cdot (\mathbf{r}_i \times \mathbf{F}_i) $$
If one assumes that internal forces do no work, then using the work energy theorem, we have $P=\frac{d}{dt} K$, which yields,
$$ \frac{d}{dt} \mathbf{L} \cdot \mathbf{\omega} = \sum \mathbf{\omega} \cdot (\mathbf{r}_i \times \mathbf{F}_i), $$
which implies the equation of motion,
$$ \frac{d}{dt} \mathbf{L}  = \sum \mathbf{r}_i \times \mathbf{F}_i. $$
Once again we see that the rigidity constraint, which guarantees the existence of $\mathbf{\omega}$, leads to the usual definitions of angular momentum and torque.
Lifting the constraint of not having any point stationary is not too difficult either, and leads to the usual torque equations again.
I recommend Feynman lectures to appreciate the energy conservation principle if you haven't seen it already.
A: The answer is because we can slide a force vector along the line of action and it does not change the system. Hence the location of a force along the line of action is not important and there only thing that matters is the moment arm of force.
Consider a force vector $\boldsymbol{F}$ acting through a point located by the position vector $\boldsymbol{r}$. Now the line of action of the force has direction $\boldsymbol{\hat{e}} = \boldsymbol{F} / \| \boldsymbol{F} \| $ and goes through $\boldsymbol{r}$. Finally, the force-moment (torque) is calculated by $$\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \tag{1} $$
Now slide the force along the line of action to a location $\boldsymbol{r}' = \boldsymbol{r} + \lambda\,\boldsymbol{\hat{e}}$. Notice that $$\require{cancel} \boldsymbol{\tau}' = \boldsymbol{r}' \times \boldsymbol{F} = (\boldsymbol{r} + \lambda \boldsymbol{\hat{e}}) \times \boldsymbol{F} = \boldsymbol{r} \times \boldsymbol{F} + \lambda\,( \cancel{ \boldsymbol{\hat{e}} \times \boldsymbol{F} })= \boldsymbol{\tau} \tag{2}$$
The idea is that the cross product $\text{(position)} \times \text{(vector)}$ only uses the perpendicular components of position for the calculation. The same applies for other quantities as well:
$$ \matrix{ \text{quantity} & \text{derivation} & \text{description} \\
\hline
\text{linear velocity} & \boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega} & \text{moment of rotation} \\
\text{angular momentum} & \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} & \text{moment of momentum} \\
\text{torque} & \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} & \text{moment of force} \\
} \tag{3}$$ 
As a mnemonic you can remember that $ \boldsymbol{r} \times \rightarrow \text{(moment of)} $ 
The real deep answer is that torque as a quantity is vector field. It changes value depending on location. The line of action of a force is fixed in space, and depending on where we measure torque about the value is $\boldsymbol{r} \times \boldsymbol{F}$ where $\boldsymbol{r}$ is the position vector of any point along the line of action.
The same can be said for linear velocity and angular momentum. Those two are vector fields of the same kind and hence use the same transformation laws as seen in (3).

You should read this answer to get more details on the nature of torque, and the geometry of mechanics.
A: Because each point in a rotating object is going in a different direction, the vector representing angular velocity was chosen to be along the axis of rotation, the only direction which characterizes the system as a whole.  Torque as a vector is defined along the axis for the same reason and because we want the torque vector to be in the same direction as the angular acceleration vector. Defining torque as the work per unit angle of rotation that could be done by a force which is tending to cause a rotation, the torque is proportional to, r, and the force component which is  perpendicular to, r.  So we need to relate three vectors (force, radius, and torque) each of which is at right angles to the others.  I'm guessing that the cross product was invented to do that.
