$SO(3)$ vs 3-Torus ${(S_1)}^3$ From rigid body rotations point of view, why are $SO(3)$ and 3-Torus not the same. Every rigid rotation is rotation about three axes. So how come $SO(3)$ is not ${(S_1)}^3$?  It seems it should be. Is it because the rotations for $SO(3)$ are somehow constrained? 
 A: Good question.  I'll try to answer it from a few perspectives, starting with the simplest (but hand-waviest), and moving to the more complicated (but rigorous).
A simpler analogous space
You probably already know that you can map the sphere $S^2$ with spherical coordinates $(\theta, \phi)$ – basically latitude and longitude.  But these are bad coordinates of the sphere; you get singularities at the north and south poles.  Well, it turns out that $\theta$ and $\phi$ are actually good coordinates for a cylinder: $\theta \in [0, \pi]$ is just the height in the cylinder, and $\phi \in [0, 2\pi)$ is the angle around the cylinder.  But then you have to map this cylinder onto the sphere.  And to do that, you basically have to pinch the top and bottom of the cylinder down to points (the north and south pole).
Now, taking a step back, we can see that this is kind of like your situation.  Imagine we wonder why $S^1 \times S^1$ is not the same as $S^2$.  Well, you can map $S^1 \times S^1$ onto $S^2$ as follows.  $S^1 \times S^1$ is a torus.  Squeeze the walls of the torus in on each other until you have a cylinder.  Then pinch the top and bottom of the cylinder down to points.
But you should feel like you've done something irreversible here.  All these squeezing and pinching operations actually change the structure of the space you're dealing with.
(As Selene points out in the comments, what we've done here is known as a suspension in topology, where this is one of the classic ways to form a new topological space from a simpler one.  And this idea of squeezing/pinching/squashing is known as taking the quotient.)
Intuitive argument
Your intuition that there are "constraints" to $SO(3)$ is correct.  To be more precise, you can actually get to $SO(3)$ from $S^1 \times S^1 \times S^1$, but only by "identifying" sets of points in the latter.  By "identify", we mean make two points the same point.
The first step is to take one of your circles $S^1$ and identify points across from each other to give you just an interval $I$.  I may not be explaining this will, but it's a simple operation that you can visualize easily.  Just take a circle in the plane, and then shrink its sides down so that the $x$ coordinate goes to zero (squash its sides in).
Now, you basically have the space of Euler angles.  This squashing operation is what makes the middle one $\beta$ only range in $[0, \pi]$, unlike the other two angles which are really circles with coordinates in $[0, 2\pi)$.  But we know that Euler angles are bad coordinates for $SO(3)$.  In particular, we have gimbal lock.  So to get to the actual space of $SO(3)$, you need to look at those coordinate singularities, which are at the ends of your interval $I$.  Each end looks like $S^1 \times S^1$.  But they're really just $S^1$, so you have to collapse those down, too.
Again, all these identifications are irreversible, so you're really changing the topology of your space.  Basically, you can map $(S^1)^3$ onto $SO(3)$, but not in a nice one-to-one way; you've really changed the space.
Purely mathematical perspective
From a formal point of view, you can prove that they're different spaces pretty easily by looking at either their group properties (as pointed out by ACuriousMind) or their topological properties.
Group theory
The direct product of groups is defined pretty simply, much like the Cartesian product of the two sets, but then define the group product to be the original group products acting on their own.  Since the circle group $U(1)$ (which is what you really meant by $S^1$) is commutative, the group $U(1) \times U(1)$ is also commutative, and therefore the whole thing $U(1) \times U(1) \times U(1)$ is commutative.  But you probably also know that $SO(3)$ is not commutative.
Topology
You can also see that these points are different by looking at their fundamental groups.  We have $\pi_1(S^1) \approx \mathbb{Z}$, and from the product property of fundamental groups, this means $\pi_1(S^1 \times S^1 \times S^1) \approx \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \approx \mathbb{Z}^3$.  Also, since the real projective space $\mathbf{RP}^3$ is topologically the same as $SO(3)$, we have $\pi_1(SO(3)) \approx \pi_1(\mathbf{RP}^3) \approx \mathbb{Z}_2$, which is the cyclic group with just two elements.  Since the fundamental group is a topological invariant, this proves that the two spaces are topologically distinct.
A: Each circle-group $S_1$ is a rotation in $\mathbb R^2$. Because rotations in the x-y-, x-z- and y-z-plane generate SO(3) one could think that if one gets a single copy for each of those plane rotations, this will result in SO(3). However, in SO(3) the order of those generating rotations is important, while in $S_1^3$ you don't define the exact order of rotation.
You just say: "Rotate in x-y, x-z and y-z"
while in SO(3) you say for instance: "Rotate in x-y then in the new x-z-plane then in the new y-z-plane"
changing the order will change the outcome.
A: A group being generated by some of its subgroups, even having trivial intersections (which is not the case in your example), doesn't allow to draw very general conclusions about its product structure if you don't take into account their commutation relations. 
The simplest example is the group of the symmetries of an equilateral triangle: this is generated by the subgroup $S$ of reflections about one fixed axis of symmetry, and the subgroup $R$ of rotations. However, it is not isomorphic to $S\times R$, the product of two cyclic groups, as you can easily check because a nontrivial reflection and a nontrivial rotation don't commute. 
It is a theorem that a group generated by commuting subgroups with trivial intersection is a direct product. When the subgroups don't commute this can to some extent be generalized to semidirect products.
