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Well, not an answer, but with probability $2/e$ the two permutations map some $i$ to the (same) $j,$ which means that both the determinant and the permanent of the difference is $0.$ Also with probability $2/e$ (not independent of the previous) $\sigma_1(i) = j, \sigma_2(j) = i,$ so again, both determinant and permanent are zero. So, the distributions will be highly atomic at $0,$ not sure about the rest of the distribution. In the sum case, in the second case ($\sigma_2^{-1} \sigma_1$ has a fixed point), the determinant is zero).