Let $\sigma\in S_n$ be a derangement. The question asks for the number of derangements $\tau,\rho$ such that $\sigma\tau\rho=1$. Fix $\tau', \rho'\in S_n$. The number of pairs $\tau,\rho$ with $\sigma\tau\rho=1$, $\tau$ conjugate to $\tau'$, and $\rho$ conjugate to $\rho'$, is
\begin{equation}
\frac{1}{n!}\lvert C(\tau')\rvert\cdot\lvert C(\rho')\rvert\sum_{\chi}\frac{\chi(\sigma)\chi(\tau')\chi(\rho')}{\chi(1)},
\end{equation}
where $\chi$ runs through the irreducible characters of $S_n$, and $C(\beta)$ is the conjugacy class of $\beta$. This formula is well known, maybe most prominently in the context of the rigidity criterion of inverse Galois theory.

All we need to do now is to let $\tau'$ and $\rho'$ run through the conjugacy classes representatives of the derangements, and some up the terms.

So the number the OP asks for is
\begin{equation}
 \frac{1}{n!}\sum_{\chi}\frac{\chi(\sigma)}{\chi(1)}\left(\sum_{\tau'}\chi(\tau')\lvert C(\tau')\rvert\right)^2,
\end{equation}
where $\tau'$ runs through the representatives of the derangements.

I don't know if this expression can be simplified. Note that if $\chi\ne1$, then $\sum_{\beta\in S_n}\chi(\beta)=0$ by orthogonality. So we may also let run $\tau'$ through the representatives of elements with at least one fixed point.

As is well known, the conjugacy classes of $S_n$ are parametrized by the partitions of $n$, and so are the irreducible characters. With regard to this parametrization, the character values can be computed, see e.g. this nice [exposition][1] (without proofs).


  [1]: http://mathematicalgemstones.wordpress.com/2012/05/21/characters-of-the-symmetric-group/