The average determinant of all integer matrices with coefficients $0,1,2$

Question

Let $S$ denote the set of $A \in M(n,\mathbb R)$ such that every entry of $A$ is either of $0$, $1$ or $2$, then is it true that $$\sum_{A \in S} \det A \ge 3^{n^2}\ ?$$

Answer 1

No. Divide $S$ into three piles: $N, Z, P$, with negative, zero, and positive determinants. Then the sum breaks into three terms. The sum for the zero-det pile is zero. And since (for $n \ge 2$) the map that swaps the first two rows of a matrix is a bijection from $N$ to $P$, the sum of $N$ and the sum over $P$ are additive inverses of each other, and therefore the total sum of the determinants is zero.

If, however, you asked about the sum of the absolute values of the dets, that'd be an intriguing question.

Post-comment addition: For $ n = 1$, the claim is true.