The name of the tensor you're looking for is the Levi-Civita or permutation tensor. In cartesian coordinates, $\epsilon_{ijk}$ is equal to $+1$ for any even permutation of 123 and $-1$ for any odd permutation.
The permutation tensor represents a directed unit volume. As such volumes have different expressions in other coordinate systems, the Levi-Civita tensor will depend on the coordinate system used. However, it represents a fundamental, coordinate-system invariant object called the pseudoscalar of the space. And there are ways to use the pesudoscalar without resorting to the tensor approach, avoiding the problem of how the pseudoscalar has different expressions in different coordinates.
The pseudoscalar enforces the notion of duality. In 3d space, vectors are dual (orthogonal to, or normal to) planes. In some circles, we call directed planes bivectors. Using the pseudoscalar in this way converts back and forth between vectors and their dual bivectors. The skew-symmetric rank-2 tensor (matrix) you have here is the direct representation of such a bivector.
You can use physical ideas to help construct a simple counterexample to show that unless additional conditions are imposed, like positive definiteness etc, the symmetric part can represent quite different things.
Consider (for simplicity) a 2D body (a unit square with one vertex at the origin) in a fixed cartesian basis, and subjected to the following homogeneous deformation.
$$
x = X + 2Y \\
y = X - Y
$$
This is not change of basis, but an actual change in the position / shape of the body, with lowercase denoting the "deformed" configuration and uppercase the coordinates of a point in the body in its original configuration.
In matrix terms
$$
\begin{bmatrix} x \\ y \end{bmatrix}
=
\begin{bmatrix}
1 & 2 \\
1 & -1 \\
\end{bmatrix}
\begin{bmatrix} X \\ Y \end{bmatrix}
$$
We'll call this matrix $F$ and write it as a sum of symmetric and skew-symmetric components
$$
F = F_{s} + F_{skew} \\
F = \frac{1}{2}(F + F^T) + \frac{1}{2}(F - F^T) \\
=
\begin{bmatrix}
1 & 3/2 \\
3/2 & -1 \\
\end{bmatrix}
+
\begin{bmatrix}
0 & 1/2 \\
-1/2 & 0 \\
\end{bmatrix}
$$
Now consider what the first (symmetric) matrix $F_s$ does to our unit square with vertices $a=(0,0), b=(1,0), c=(1,1) \; and \; d=(0,1)$ in that order. (We might as well start with symmetric F, but anyway).
The new coordinates of the unit square are (let's call them $a', b', c', d')$ under this symmetric map are
\begin{bmatrix}
a' & b' & c' & d' \\
0 & 1 & 5/2 & 3/2 \\
0 & 3/2 & 1/2 & -1 \\
\end{bmatrix}
Looks ok. However, if you sketch this figure you'll see that the handedness of the figure has flipped. In particular, its Jacobian is negative, so the area changes sign. What this implies is that no matter how you parametrize or decompose (say using products) the map $F_s$, at some stage, material in the square must interpenetrate to reach the final configuration. So in this instance, the symmetric matrix with no other restriction represents an impossible deformation (if you look at it physically). Mathematically, it represents a change of handedness by interchanging sides PLUS a stretch in this case.
Even weirder things can happen when the jacobian is 0. The symmetric map
\begin{bmatrix}
1 & -1 \\
-1 & 1 \\
\end{bmatrix}
results in collapsing our unit square into a straight line passing through (-1,1) and (1,-1) with obviously no area! So you see that symmetry by itself doesn't tell you what's going on geometrically unless you look at additional characteristics of the map.
Best Answer
Note that $Sv=0$.
In other words $v$ resides in the nullspace of $S\,-$ in fact it's the only vector in the space.
With this insight, we can write a formula in terms of the pseudoinverse and an arbitrary vector $$\eqalign{ b &= (I-S^+S)\,a \cr v &= b \,\frac{\|v\|}{\|b\|}\cr }$$ There is an explicit formula for the pseudoinverse of a $3\times 3$ skew-symmetric matrix $$\eqalign{ S^+ &= \frac{-2S}{\|S\|_F^2}\cr }$$ Also, the length of $v$ is related to the norm of $S$ $$\eqalign{ \|S\|_F^2 &= 2\,\|v\|^2 \implies \|v\| &= \frac{\,\|S\|_F}{{\sqrt 2}} \cr }$$ Assembling all these pieces yields $$\eqalign{ b &= \|S\|_F^2\,a + 2S^2a \cr v &= \pm\,b\,\,\frac{\|S\|_F}{{\sqrt 2}\,\,\|b\|_F} \cr }$$