I've addressed this problemsmall cases with MILPlinear programming. For $n\leq 5$, the maximum happen to be just $n$, but for $n=6$ it's only $\frac{16}3$. Here are the constructed solutions:
Max: 3
[ 0 -1 1]
[-1 1 0]
[ 1 0 -1]
Max: 4
[ 1/4 -1/4 -3/4 3/4]
[ 3/4 1/4 -1/4 -3/4]
[-3/4 3/4 1/4 -1/4]
[-1/4 -3/4 3/4 1/4]
Max: 5
[ 1/2 1/2 -1/2 -1/2 0]
[ 0 -1/2 -1/2 1/2 1/2]
[-1/2 0 1/2 -1/2 1/2]
[-1/2 1/2 0 1/2 -1/2]
[ 1/2 -1/2 1/2 0 -1/2]
Max: 16/3
[ -3/8 -1/3 3/8 11/24 -7/24 1/6]
[11/24 -1/3 -7/24 -3/8 3/8 1/6]
[ 3/8 1/6 11/24 -7/24 -3/8 -1/3]
[-7/24 1/6 -3/8 3/8 11/24 -1/3]
[ -1/3 1/6 1/6 -1/3 1/6 1/6]
[ 1/6 1/6 -1/3 1/6 -1/3 1/6]