For an $n$-by-$n$ unitary matrix $U$, what's the minimal value of the real part of $\Delta(U)=\det(U^*)\prod_i U_{ii}$?
Let $V$ be the orthogonal matrix with diagonal entries equal to $1-2/n$ and all other entries equal to $-2/n$. This achieves $\Delta(V)=-(1-2/n)^n$, which computer experiments suggest is optimal. Interestingly this would mean that the large $n$ limit is $-e^{-2}$.
For $n=2$ the minimum is $0$, which can be proven by writing $U$ in the form
$$\begin{pmatrix}\alpha & \beta \\ -e^{-i\theta}\bar\beta & e^{-i\theta}\bar\alpha\end{pmatrix}.$$
The average value of $\Delta(U)$ across the unitary group is $1/n!$. Indeed, for any permutation $\sigma$ with permutation matrix $P_\sigma$, $\Delta_\sigma(U)=(-1)^\sigma\det(U^*)\prod_i U_{i,\sigma(i)}$ equals $\Delta(UP_\sigma)$. The sum $\sum_\sigma \Delta_\sigma(U)$ equals $\det(U^*)\det(U)=1$, and each $\int_{U(n)}\Delta_\sigma(U)dU$ is equal because multiplication by $P_\sigma$ preserves the Haar measure.
Best Answer
We consider the real case.
$\textbf{Proposition}$. Let $n>2$ and
$f:U=[u_{i,j}]\in O(n)\mapsto \det(U^T)\Pi_{i=1}^n u_{i,i}$.
Then the minimum of $f$ is $m=-(1-2/n)^n$.
$\textbf{Proof}$. Since $O(n)$ is compact, the lower bound of $f$ is reached in at least a matrix $A=[a_{i,j}]\in O(n)$. Note that if we change a column of $U\in O(n)$ into its opposite, then the obtained matrix $U'\in O(n)$ satisfies $f(U)=f(U')$.
Consequently, we may assume that, for every $i$, $a_{i,i}\geq 0$. Since we know, from the OP, that $m\leq -(1-2/n)^n< 0$, we deduce that, for every $i$, $a_{i,i}>0$ and $\det(A)<0$.
Then $-1\in spectrum(A)$ and $\sum_i a_{i,i}=trace(A)\leq n-2$.
Consequently, $0<\Pi_i a_{i,i}\leq (\dfrac{n-2}{n})^n$ (fixing the sum of the $(a_i)$ in $n-2$, the max of the product is reached when the $(a_{i,i})$ are equal)
and we are done. $\square$