Solution:
Answer: $15(2013!)$

First, note that
$$\sum_{\sigma \in S} f(\sigma)^4 = \sum_{\sigma \in S} \sum_{1 \leq a_1, a_2, a_3, a_4 \leq 2013} g\left(\sigma, a_1, a_2, a_3, a_4\right)$$
where $g\left(\sigma, a_1, a_2, a_3, a_4\right) = 1$ if all $a_i$ are fixed points of $\sigma$ and $0$ otherwise. (The $a_i$'s need not be distinct.) Switching the order of summation, we find that the desired sum is
$$\sum_{1 \leq a_1, a_2, a_3, a_4 \leq 2013} \sum_{\sigma \in S} g\left(\sigma, a_1, a_2, a_3, a_4\right)$$
Note that the inner sum is equal to the number of permutations on $\{1,2, \ldots, 2013\}$ that fix $a_1, a_2, a_3$, and $a_4$. This depends on the number of distinct values the $a_i$ take. If they take on exactly $k$ distinct values, then the inner sum will evaluate to $(2013-k)!$, because $\sigma$ can be any permutation of the remaining $2013-k$ elements. (For example, if $a_1 = a_2$ but $a_1, a_3$, and $a_4$ are distinct, the inner sum is $2010!$ because $\sigma$ can be any permutation that fixes $a_1, a_3$, and $a_4$.)

Now, suppose we are given which of the $a_i$ are equal (for example, we could be given $a_1 = a_2$ but $a_1, a_3, a_4$ mutually distinct, as per the above example). Assuming there are $k$ distinct values among the $a_i$, there are $2013(2013-1) \cdots (2013-k+1)$ ways to choose the $a_i$. At this point, there are $(2013-k)!$ ways to choose $\sigma$ on the remaining $(2013-k)$ values such that it fixes the $a_i$, for a total of $2013!$ choices for $(\sigma, a_1, a_2, a_3, a_4)$ such that $g(\sigma, a_1, a_2, a_3, a_4) = 1$ and the $a_i$ satisfy the correct equality relations.

Thus the answer is $2013!$ times the number of ways to choose equivalence classes on the $a_i$, so the problem reduces to finding the number of ways to partition $4$ elements into nonempty sets. This process can be accelerated by doing casework based on the number of sets:

a. One set must contain all four elements, only one possibility. (i.e. all the $a_i$ are equal)

b. Either one set contains $3$ elements and the other contains the fourth ($4$ possibilities) or one set contains $2$ elements and the other contains the other two ($3$ possibilities). (i.e. there are two distinct values of $a_i$)

c. One set contains two elements, the other two each contain one. There are $\binom{4}{2} = 6$ ways to choose the two elements in the set with two elements, and this uniquely determines the partition. (i.e. there are three distinct values of $a_i$)

d. All sets contain one element, in which case there is only one possibility. (i.e. all the $a_i$ are distinct)

Thus the number of ways to construct such a partition is $1 + 4 + 3 + 6 + 1 = 15$, and our answer is $15 \cdot 2013!$.