I don't know that it is of the type you wish, but there is a formula of sorts.

Consider the more general case of a finite group $G$ acting diagonally by conjugation on the set of $k$-tuples of elements of $G$.  For a fixed $g$ in $G$, the number of fixed points of $g$ is $|C_G(g)|^k$.  Let $g_1,\dots,g_c$ be a set of representatives for the conjugacy classes of $G$.  Applying Burnside's Lemma and grouping together elements in the same conjugacy class, we see that the number of orbits of $G$ in the given action is the sum over all $g_i$ of $|C_G(g_i)|^{k-1}$.

So, in your case, we have the sum of $|C_G(g_i)|$ over a set of representatives for the conjugacy classes of $S_n$.  As you noted, these classes are parameterized by partitions of n, and if such a partition $p$ has $a_j$ parts of size $j$ for each $j$ in $[n]$, the order of the corresponding centralizer is the product over all such $j$ of $(a_j)!j^{a_j}$.  Thus we get the sum over all partitions of n of such products.

Maybe it is worth remarking that, for any $G$, when $k=2$ the number in question is the sum of the square norms of the entries of the character table of $G$.