If we use the heuristics that a number $n$ is prime with probability $1 / \log n$, the expected number of $m \le x$ with $f(m)$ prime is $\sim \log \log x$. I don't think checking for $m \le 1300$ is good enough support.

On the other hand, I think a naive probabilistic argument gives an example of $\lvert A \rvert + \lvert B \rvert = O(\sqrt{n \log n})$.

I personally doubt this could be true. Can you find an example of $\lvert A \rvert + \lvert B \rvert = 11$ for $G = (\mathbb{Z}/3)^3$?

A unit triangle can cover more than $1$ of the circumference! I think it can cover up to $2/\sqrt{3}$.

It seems that, computationally, there is no orthogonal period, but someone who knows number theory would need to confirm this.

Does this upload.wikimedia.org/wikipedia/commons/6/6a/2-uniform_17.png have an orthogonal period?

I'm not an expert in analysis, but I think this kind of argument works: Suppose $f$ is bounded by a polynomial. The Fourier transform $\hat{f} : T^n \to \mathbb{R}$ is then a tempered distribution. From the harmonic condition, we get $\hat{f} \cdot (\sum_k (e^{2 \pi i x_k} + e^{- 2 \pi i x_k}) - 2d) = 0$ and so $\hat{f}$ is supported on the origin. This implies that $\hat{f}$ is a linear combination of $D^\alpha \delta$ and thus $f$ is indeed a polynomial.

The answer seems to be related to the covering density of the set $\{1, \dots, n\}$ in the multiplicative group generated by $1, \dots, n$ (viewed as a lattice).