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 $s_1$.