It is a long comment to show that problem in arbitrary dimension is far from trivial.
If we set simplified case: $B = I$, then:
$$
n = 2: \det(I + A) = 1 + \det(A) + Tr(A)
$$$$
n = 3: \det(I + A) = 1 + \det(A) + Tr(A) + \frac{Tr^2(A) - Tr(A^2)}{2}
$$$$
n = 4: \det(I + A) = 1 + \det(A) + Tr(A) + \frac{Tr^2(A) - Tr(A^2)}{2} + \frac{Tr^3(A) - 3Tr(A)Tr(A^2) + 2Tr(A^3)}{6}
$$
General sequence comes from relation between determinant and exponent of the trace of the matrix logarithm.
I doubt that there is a trivial solution in general case.
IT IS NOT AN ANSWER.
It is a long comment to show that problem in arbitrary dimension is far from trivial. If we set simplified case: $B = I$, then: $$ n = 2: \det(I + A) = 1 + \det(A) + Tr(A) $$$$ n = 3: \det(I + A) = 1 + \det(A) + Tr(A) + \frac{Tr^2(A) - Tr(A^2)}{2} $$$$ n = 4: \det(I + A) = 1 + \det(A) + Tr(A) + \frac{Tr^2(A) - Tr(A^2)}{2} + \frac{Tr^3(A) - 3Tr(A)Tr(A^2) + 2Tr(A^3)}{6} $$ General sequence comes from relation between determinant and exponent of the trace of the matrix logarithm. I doubt that there is a trivial solution in general case.