Can any rank tensor be decomposed into symmetric and anti-symmetric parts? I know that rank 2 tensors can be decomposed as such. But I would like to know if this is possible for any rank tensors?
 A: A (higher) $n$-rank tensor $T^{\mu_1\ldots \mu_n}$ with $n\geq 3$ cannot always be decomposed into just a totally symmetric and a totally antisymmetric piece.
In general, there will also be components of mixed symmetry.
The symmetric group $S_n$ acts on the indices 
$$(\mu_1,\ldots ,\mu_n)\quad \longrightarrow\quad (\mu_{\pi(1)},\ldots ,\mu_{\pi(n)})$$ 
via permutations $\pi\in S_n$. One can decompose the tensor $T^{\mu_1\ldots \mu_n}$ according to irreps (irreducible representations) of the symmetric group. 
Each irrep corresponds to a Young tableau of $n$ boxes. For instance, a single horizontal row of $n$ boxes corresponds to a totally symmetric tensor, while a single vertical column of $n$ boxes corresponds to a totally antisymmetric tensor. But there are also other Young tableaux with a (kind of) mixed symmetry.
Here is a Google search for further reading.
A: No.
Definitions:

*

*A rank-n tensor is a linear map from n vectors to a scalar.

*A symmetric tensor is one in which the order of the arguments
doesn't matter.

*An antisymmetric tensor is one in which transposing two arguments multiplies the result by -1.

Suppose we have some rank-3 tensor $T$ with symmetric part $S$ and anti-symmetric part $A$ so
$$T(a,b,c) = S(a,b,c) + A(a,b,c)$$
where $a,b,c\,$ are arbitrary vectors. Transposing $c$ and $a$ on the right hand side, then transposing $a$ and $b$, we have
$$T(a,b,c) = S(c,a,b) + A(c,a,b)$$
so we conclude
$$T(a,b,c) = T(c,a,b)$$
it is trivial to construct a counterexample, so not all rank-three tensors can be decomposed into symmetric and anti-symmetric parts.
