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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 $\sigma_1\sigma_2^{-1}$. Distributions of cycle statistics like these are easy to obtain with the exponential formula.