This is certainly a well established topic. Probably originating at the NIST (which would then have been called the National Bureau of Standards.)

A [Weighing Matrix][1] of weight $w$ and order $n$ is an $n \times n$ $0,\pm1$ matrix $A$ with $A^tA=wI_n.$ The case $w=n$ is a [Hadamard Matrix][2] with $A^tA=nI_n.$ Other than $n=1,2$ this requires $n=4m.$ It is an open question if they exist for every such $n.$ According to the linked article $668, 716, 892, 1004, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948,$ and  $1964$ are the only multiples of $4$ up to $2000$ for which such a design is unknown. 

So $\sigma^2_n=\frac{\sigma^2}{n}$ for $n$ such that there is a Hadamard matrix of order $n.$

I would have guessed that for such an $n$ the optimal thing (in whatever sense) to do for $n-1$ is take a Hadamard matrix and delete a row and column. You showed that in some sense that is not optimal for $4-1=3.$

I really wonder what the case is for $n=8-1=7.$ The underlying design is a Fano Plane which seems as if it must be optimal (yet it isn't). For example: $$\left[ \begin {array}{ccccccc} -1&-1&1&-1&1&1&1\\  1
&-1&-1&1&-1&1&1\\  1&1&-1&-1&1&-1&1
\\  1&1&1&-1&-1&1&-1\\  -1&1&1&1&-1&
-1&1\\  1&-1&1&1&1&-1&-1\\  -1&1&-1&
1&1&1&-1\end {array} \right] 
$$

Here $AA^t=7I-(J-I)$ is all $-1$ except $7$'s on the diagonal. The inverse of this is $\frac14I+\frac18(J-I).$

If the top left entry is changed to $0$ then the maximum entry on the diagonal is still $\frac14$  but in four of the seven diagonal positions one has $\frac{13}{64}.$  So this is, in some sense, even betterI didn't experiment with going further than that.


  [1]: https://en.wikipedia.org/wiki/Weighing_matrix
  [2]: https://en.wikipedia.org/wiki/Hadamard_matrix