(Something to start with.)

Denote the permanent by $P$. We have ${\mathbb E}(P)=(1-p)^nn!$. Now look at ${\mathbb E}(P^2)$. This is a sum over all pairs of permutations $\pi,\sigma$ of $(1-p)^{2n-fix(\pi^{-1}\sigma)}$, where $fix(\tau)$ denotes the number of fixed points of a permutation $\tau$. Thus $${\mathbb E}(P^2)=n!(1-p)^{2n}\sum_{\tau\in S_n} (1-p)^{-fix(\tau)}=n!^2(1-p)^{2n}\sum_{i=0}^n\frac1{i!}\left(\frac{p}{1-p}\right)^i,$$
see [here][1], for example. For small $p$ this is close to $({\mathbb E} P)^2$ and thus we have a sort of concentration. When $p$ grows, the ratio ${\mathbb E}(P^2)/({\mathbb E}(P))^2$ grows, so we lose concentration.

  [1]: http://www.combinatorics.org/ojs/index.php/eljc/article/viewFile/v20i2p26/pdf