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Using standard symmetric function notation, we have \begin{eqnarray*} \sum_{n\geq 0}\sum_{\lambda,\mu\vdash n} \frac{1}{n!}\left(\sum_{\pi\in D_n}\chi_\lambda(\pi)\chi_\mu(\pi)\right) s_\lambda(x)s_\mu(y) & = & \sum_{n\geq 0}\frac{1}{n!} \sum_{\pi\in S_n}\left.p_{\rho(\pi)}(x)p_{\rho(\pi)}(y)\right|_{p_1(x)=0}\\ & = & \sum_\nu \left.s_\nu(x)s_\nu(y)\right|_{p_1(x)=0}\\ & = & e^{-p_1(x)p_1(y)}\sum_\nu s_\nu(x)s_\nu(y), \end{eqnarray*} since $$\sum_\nu s_\nu(x)s_\nu(y)=\exp \sum_{m\geq 1}\frac {p_m(x)p_m(y)}{m}. $$ Thus your sum is obtained by expanding $e^{-p_1(x)p_1(y)}\sum_\nu s_\nu(x)s_\nu(y)$ in terms of Schur functions and taking $n!$ times the coefficient of $s_\lambda(x)s_\mu(y)$. To do this expansion you could write $$ e^{-p_1(x)p_1(y)} = \sum_{m\geq 0} (-1)^m\frac{s_1(x)^ms_1(y)^m}{m!} $$ and iteratively apply Pieri's formula for multiplying a Schur function by $s_1$.