Here's the first few $n$ with the maximum average distance $m$ over all corner configurations, and a configuration which realizes it. I've changed the interval from $[-1,1]$ to $[0,1]$ to make it easier to read.

\begin{align}
n=2, & \ m=\sqrt{2}\simeq 1.414 & (0,0)\\
& & (1,1)\\
n=3, & \ m=\sqrt{2}\simeq 1.414 & (0,0,1)\\
& & (0,1,0)\\
& & (1,0,0)\\
n=4, & \ m=\frac{\sqrt{2}+2\sqrt{3}}{3}\simeq 1.626 & (0,0,0,0)\\
& & (0,0,1,1)\\
& & (1,1,0,1)\\
& & (1,1,1,0)\\
n=5, & \ m=\frac{\sqrt{2}+3\sqrt{3}+\sqrt{4}}{5}\simeq 1.722 & (0,0,0,0,0) \\
& & (0,0,0,1,1)\\
& & (0,1,1,0,0)\\
& & (1,0,1,0,1)\\
& & (1,1,0,1,0)\\
\end{align}
This suggests that choosing all coordinates independently and randomly may get an average close to the maximum, and for large $n$ that average is roughly
$$\left(1-\frac{1}{8n}\right)\sqrt{\frac{n}{2}}.$$