Why is the anti-symmetric tensor more important than symmetric tensors? In differential geometry, the differential forms are anti-symmetric tensors. 
So, why is the anti-symmetric tensor like 
$ d x_1 \otimes dx_2 - d x_2 \otimes d x_1 $, more important than the symmetric tensor like $ d x_1 \otimes dx_2 + d x_2 \otimes d x_1 $ for the theory?
Why do we need anti-symmetric tensors for the exterior algebra, the wedge?
In quantum mechanics, we have fermionic wave functions and bosonic wave functions. Is the former more fundamental than the latter in a certain sense?
 A: Your question touches on many aspects of mathematics and physics at once - here is how I see it:
In (linear) algebra, the exterior algebra arises (form one viewpoint) solely through the need to define the determinant in a non-basis-dependent way, since it is, up to normalization, simply the sole basis element of the top exterior power. Since the determinant for matrices who possess two linearly dependent columns is zero, $x \wedge x = 0$ is the way to enforce this, and over fields with characteristic not 2, this implies antisymmetry. The actual way we construct this exterior power is by dividing the ideal generated by $I := \{x\otimes x | x \in V\}$ out of the tensor algebra over a vector space $V$, so we can see the exterior power as consisting of antisymmetric tensors, although sometimes one should remember that it really is a quotient of the tensors. In this way, antisymmetry of the exterior power arises naturally.
Now, in differential geometry, the differential forms arise through the cochain complex of the de Rham cohomology, which is just the complex of $p$-forms with the exterior derivative as the coboundary. You ask: "Why do we take the degrees of the exterior power of $T_p M$ and not the degrees of the tensor algebra as defining these forms?" The answer is simple: On mere tensors, $\mathrm{d}^2 = 0$ would not hold, thus the degrees of the tensor algebra do not define a cochain complex, thus they do not generate a cohomology. If you ask why we want a cochain complex/cohomology, it is because, by virtue of Eilenberg-Steenrod, every (co)homology is intrinsically related to the geometric intuiton of loops, surfaces and hypersurfaces on a manifold.
Fermionic and bosonic functions/objects are a wholly different story: They arise through different representations of the Lorentz group $\mathrm{SO}(1,3)$, or rather, its double cover $\mathrm{Spin}(1,3)$. Since we want Lorentz invariance, the Lorentz group must have a way to act on our fields/objects. It can only do so if they are in a representation of our symmetry group, and when we construct reps, we generally do so by representing only the infinitesimal generators of the symmetries, which lie in the Lie algebra - but this is the same for the Lorentz group and its double cover, and representing the algebra will always induce reps of the universal cover, so we are stuck with the reps of $\mathrm{Spin}(1,3)$, and this has representations with "spin" eigenvalue $j \in \mathbb{N}$ and $j \in \mathbb{N}+\frac{1}{2}$. Fields transforming in one of the former reps are bosonic, fields transforming in one of the latter are fermionic.
A: In differential geometry we want to define geometrical objects out of objects of lower dimension:  areas out of line segments,  volumes out of an area segment and a line segment, etc.  For areas, think of the parallelogram formed by two line segments.  We should have it that the area formed by two identical line segments be zero.  We can insure this happens by using the antisymmetric combination.  In fact, in the Euclidean plane, the antisymmetric product of the coordinates of the ends of the line segments does indeed give the area. (Assuming one end of each segment is at the origin.)  Similarly, the three dimensional volume defined by three line segments is equal to an antisymmetric product of the coordinates of the ends of the line segments, although the expression is more complicated than it is for areas.
