Consider $ S(k,e)=\sum_{i=0}^{k} {e \choose i} (1-p)^{e-i} p^i. $ Then $S(k,e)=1$ if $k\ge e$, and your sum $S_{n,e}$ is $ S_{n,e}=(n-1)S(n-2,e)-epS(n-3,e-1). $ Hence $S_{n,e}=n-1-ep$ for every $n\ge e+2$.
If $n, there might not exist so simple expressions of $S_{n,e}$ in all generality. Note however for small $n$ that $S(0,e)=(1-p)^e$, hence $S_{0,e}=S_{1,e}=0$ and $S_{2,e}=(1-p)^e$.
Edit According to E.C. Molina (Application to the Binomial Summation of a Laplacian Method for the Evaluation of Definite Integrals, Bell System Technical Journal, v8: i1 January 1929, 99-108, available here), for $k, $ S(k,e)=B_{k,e}(p)/B_{k,e}(0),\quad\mbox{where}\ B_{k,e}(p)=\int_p^1x^{k}(1-x)^{e-k-1}\mathrm{d}x. $ The author then presents some approximations of $S(k,e)$.