Hmm, if you write out the determinant using the Leibniz formula then the terms of the sum are actually independent so the variance of the sum will be sum of the variances, each of which is 1. So the total variance will just be n!.
Are you sure the terms are independent? In the (3\times 3) case, for instance, you have the terms (a_{1,1}a_{2,2}a_{3,3}) and (-a_{1,1}a_{2,3}a_{3,2}) which are not independent...
A product of Bernoulli (mean zero) random variables still has mean zero. Hence E(det(A)]=0.
E[det(A)^2] = sum of entries with squares + sum of entries without squares. The 2nd half is zero, the first half is a sum of 1s, hence var[det(A)^2] = n!
Actually ThisGuy's argument still holds, if you replace independent with uncorrelated.
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