Hadamard's maximal determinant problem

Remarks

Hadamard's inequality gives $\abs{\det A} \le n^{n/2}$ for such matrices. Equality holds exactly when the rows of $A$ are pairwise orthogonal, that is, when $A$ is a Hadamard matrix. A Hadamard matrix of order $n$ can exist only for $n = 1$, $n = 2$ or $n$ divisible by $4$. The Hadamard conjecture, that one exists for every multiple of $4$, is open. For other orders the maximum is smaller than $n^{n/2}$, and it is known only for some values of $n$.

The determinant of an $n \times n$ matrix with entries $\pm 1$ is divisible by $2^{n-1}$. The maximum for $n$ equals $2^{n-1}$ times the maximal determinant of an $(n-1) \times (n-1)$ matrix with entries $0$ and $1$. So the problem is the same as the maximal determinant problem for $0/1$ matrices.

Small cases: for $n = 2$, the maximum is $2$, reached by $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$. For $n = 3$, the bound $3^{3/2} \approx 5.2$ and the divisibility by $4$ show that the maximum is at most $4$, and $4$ is reached.