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$G$ acting diagonally by conjugation on the set of k$k$-tuples of elements of G$G$. For a fixed g$g$ in G$G$, the number of fixed points of g$g$ is |C_G(g)|^k$|C_G(g)|^k$. Let g_1,...,g_c$g_1,\dots,g_c$ be a set of representatives for the conjugacy classes of G$G$. Applying Burnside's Lemma and grouping together elements in the same conjugacy class, we see that the number of orbits of G$G$ in the given action is the sum over all g_i$g_i$ of |C_G(g_i)|^{k-1}$|C_G(g_i)|^{k-1}$.
So, in your case, we have the sum of |C_G(g_i)|$|C_G(g_i)|$ over a set of representatives for the conjugacy classes of S_n$S_n$. As you noted, these classes are parameterized by partitions of n, and if such a partition p$p$ has a_j$a_j$ parts of size j$j$ for each j$j$ in [n]$[n]$, the order of the corresponding centralizer is the product over all such j$j$ of (a_j)!j^{a_j}$(a_j)!j^{a_j}$. Thus we get the sum over all partitions of n of such products.
Maybe it is worth remarking that, for any G$G$, when k=2$k=2$ the number in question is the sum of the square norms of the entries of the character table of G$G$.
