I will abuse notation by identifying a permutation and the matrix it represents. We can denote by $E(\sigma), O(\sigma)$ the number of even and odd cycles that $\sigma$ decomposes into. Given two permutations $\sigma_1,\sigma_2$ we can compute the following:
$$\det(\sigma_1+\sigma_2)=\left\{
	\begin{array}{ll}
		(-1)^{E(\sigma_1)}2^{O(\sigma_1\sigma_2^{-1})}  & \mbox{if } E(\sigma_1\sigma_2^{-1})=0 \\
		0 & \mbox{otherwise }
	\end{array}
\right. $$
$$\operatorname{per}(\sigma_1+\sigma_2)=2^{E(\sigma_1\sigma_2^{-1})+O(\sigma_1\sigma_2^{-1})}$$
$$\operatorname{per}(\sigma_1-\sigma_2)=\left\{
	\begin{array}{ll}
		2^{E(\sigma_1\sigma_2^{-1})}  & \mbox{if } O(\sigma_1\sigma_2^{-1})=0 \\
		0 & \mbox{otherwise }
	\end{array}
\right. $$
and trivially $\det(\sigma_1-\sigma_2)=0$ since the vector of all 1's is always in the kernel of $\sigma_1-\sigma_2$. These calculations follow from noticing that the matrices decompose as direct sums of smaller matrices corresponding to each cycle of $\sigma_1\sigma_2^{-1}$. <a href="https://en.wikipedia.org/wiki/Random_permutation_statistics">Distributions of cycle statistics</a> like these are easy to obtain with the exponential formula.