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If your matrix $A$ is close to the identity I guess that you could try the following approximation \begin{align} \log(\det(I + \epsilon B)) &= \log\det(I + \epsilon_0 B) + (\epsilon - \epsilon_0)\operatorname{tr}(B (I + \epsilon_0 B)^{-1}) - (\epsilon - \epsilon_0)^2 \operatorname{tr}(B (I + \epsilon_0 B)^{-1} B (I + \epsilon_0 B)^{-1}) + O(\epsilon^3) \...
The LU decomposition will give you what you want with only $\tfrac{2}{3}n^3 + \mathcal{O}(n^2)$ FLOPs. The linear system is solved by solving two triangular systems. The determinant is the product of the determinants of L and U, which, in turn, are the products of the diagonal elements. I should also add that we can't really say anything about the ...