A square constraint is a constraint of the type $$\{x \in \mathbb{R}| x \in [a,b], a < b\}$$ (generalization to arbitrary dimension is using Cartesian products, with $<$ overloaded) $$\{x \in \mathbb{R}^n| x \in [a,b]^n, a < b\}$$
We know that the $l_\infty$ norm of a vector $x$ has an pre-image that looks sort of like a square
$$\|x\|_\infty = \max_{i = 1, \ldots, n} |x_i|$$
Is there some connection between the $l_\infty$ ball and the square constraint (similar to how there is a relationship between the Euclidean ball and the circle constraint)?
Indeed there is a connection. For $x \in \mathbb R^n$, the $\ell_\infty$ ball $$\{x \in \mathbb R^n: \| x\|_\infty \leq a\}$$ is equivalent to saying that $|x_i| \leq a$ for every $i \in \{1, \dots, n\}$. This in turn corresponds to the square constraint $$\{x \in \mathbb R^n: x \in [-a,a]^n\}. $$