If $n$ is odd, there must be an odd number of zero terms in the inner product of any two rows. In particular, for any two rows, there are an odd number of columns where both are zero. If all the zeroes don't lie in a column, the combinatorics becomes interesting (and the problem seems hard). If they're all in a column and there exists a Hadamard matrix of order $n-1$, it seems the direct sum with a $1\times 1$ matrix would be hard to beat, at least at large orders.

If $n \equiv 2 \bmod 4$ there do not exist three orthogonal $\pm 1$ vectors. So you could have at most two $n$'s in the Gram matrix, and if you have at least one a parity argument shows the determinant would be at most the square root of $n^2(n-2)^{n-2}$. You can do better with a conference matrix, which is a matrix with zero's on the diagonal, $\pm 1$ elsewhere satisfying $MM^{\top} = (n-1)I_{n}$. So these should be the optimal solutions when $n \equiv 2 \mod 4$, when they exist. The order must be a sum of two squares; the Paley construction with $q \equiv 1 \mod 4$ gives an infinite family.

There are known to be exponentially many (in $n$) equivalence classes of Hadamard matrices at orders of the form $2^{n}$. This is a result of Eric Merchant.

Maximising the function $\nu_{n, 1}$ is equivalent to maximising $\sum_{H} |X| - 2|X \cap H|$ where $X \subseteq \mathbb{F}_{2}^{n}$, and the sum is over hyperplanes of $\mathbb{F}_{2}^{n}$. So the function is large when $X$ is 'unbalanced' with respect to many hyperplanes. Perhaps there is a construction for such objects in the finite geometry literature?

The incidence structures you mention seem to have been studied under the name of partial linear spaces. Also - in a $t$-$(v,k,\lambda)$ design, it's normal to use $\lambda$ for the number of blocks containing any $t$-set of points. When the design is symmetric, then $t = 2$ and the block intersection numbers are also all equal to $\lambda$ but this does not hold in general. I don't know if there's a better symbol than $\lambda$ for the block intersections in this context...

There's a Chapter in Marshall Hall's Combinatorial Theory where all this is laid out in detail.

Sorry - I still don't see what you're trying to say. Whiteman's paper is an elaboration of the cyclotomic construction for difference sets. This is normally applied to a group of prime order. It might help you to understand the base case thoroughly, including the relation between the cyclotomic numbers and expressions of the prime as a sum of squares. Then the material in Whiteman's paper is intricate, but nothing one would not expect. In particular, you should see why Whiteman carries out the constructions he does for $-1$, and why this is only a detail in the main argument.

I am not sure what you are asking. The given value of $m$ solves $(xy)^{m} \equiv -1 \bmod pq$ when both $\frac{p-1}{e}$ and $\frac{q-1}{e}$ are odd. Otherwise, it does not, because there is no solution to the congruence. In that case, you need to multiply by a power of $x$.

I agree that $x \equiv 8 \bmod 35$. And $y \equiv 31 \bmod 35$. Then $xy \equiv 3 \bmod 35$, and this generates a cyclic subgroup of order $12$. But the multiplicative group has order $24$, and the elements which are not powers of $3$ belong to $8\langle 3 \rangle = \{ 8 \cdot 3^j \bmod 35 \}$. You can compute that $8 \cdot 3^9 \equiv -1 \mod 35$.

First -- the multiplicative group of $\mathbb{Z}_{pq}$ is not cyclic, so doesn't have a primitive root. Compute some examples to convince yourself of this. Second -- in the third paragraph, $m$ must be even to get a solution $\bmod p$ and odd for a solution $\bmod q$, which is impossible. So no power of $xy$ evaluates to $-1 \bmod pq$. I think your confusion here will be resolved by thinking about my first sentence, and computing some explicit examples.

This looks similar in idea to the S-boxes used in AES, DES and similar cryptosystems. (The details are more complicated but they typically apply a permutation of the input bits, then apply a fixed function to small blocks, and iterate these steps for a large number of rounds.) When they were developed, there was a lot of research into necessary and sufficient conditions for security. Maybe these translate into guarantees of good pseudo-random behaviour here? At least, that's where I would start looking.

For cyclic groups, replacing addition with subtraction in the question gives a relation to graceful labellings of graphs. For cyclic groups, these exist when $n \equiv 3 \mod 4$ but not when $n \equiv 1 \mod 4$ by a theorem of Rosa. Unfortunately this doesn't help answer the question...

If the answer to your precise question is not 'yes' then some minor modification should be true, though note the orbital matrices are not always symmetric.