Summing over normalized characters of the permutation group

The quantity you are asking about is in fact a well-known expression: When multiplied by $n!$, it is the number of ordered pairs $\sigma, \tau \in S_{n}$ such that $[\sigma, \tau] = \mu$, where $[\sigma, \tau] = \sigma^{-1}\tau^{-1}\sigma \tau$ is the commutator of $\sigma$ and $\tau$. However, I do not know how to relate this to the disjoint cycle structure of $\mu$, except to say that this quantity is clearly zero if $\mu$ is an odd permutation.

This is just an observation. I normalize your problem by $n!$, to get rid of denominators.

Let $A_n(\mu) := n! \sum_{\lambda \vdash n} \chi^{\lambda}(\mu)/f^\lambda$.

Define $B_n(x) := n! \sum_{\lambda \vdash n} \frac{p_\lambda(x)}{f^\lambda}$. Then $A_n(\mu) = \langle B_n(x), s_\mu \rangle$. That is, $A_n(\mu)$ is the coefficient of $s_\mu$ when expanded in the Schur basis.

The Schur expansion of $B_n(x)$ for $n=1,2,\dotsc$ are \begin{array}{l} s_{1} \\ 4 s_{2} \\ 15 s_{3}+6 s_{21}+9 s_{111} \\ 76 s_{4}+64 s_{22}+44 s_{31}+76 s_{211}+12 s_{1111} \\ 368 s_{5}+628 s_{32}+416 s_{41}+580 s_{221}+792 s_{311}+344 s_{2111}+200 s_{11111} \end{array} Perhaps there is some pattern...