Does the Lebesgue Differentiation Theorem hold for regular polytopes?

Lebesgue's differentiation theorem states that if $$x$$ is a point in $$\mathbb{R}^n$$ and $$f:\mathbb{R}^n\rightarrow\mathbb{R}$$ is a Lebesgue integrable function, then the limit of $$\frac{\int_B f d\lambda}{\lambda(B)}$$ over all balls $$B$$ centered at $$x$$ as the diameter of $$B$$ goes to $$0$$ is equal almost everywhere to $$f(x)$$. But if you replace balls with other kinds of set with diameter going to $$0$$, this need not be true. For instance it need not be true if you replace balls with rectangles.

It does hold for squares, though, and more generally for $$n$$-dimensional cubes. So my question is, is it known if it holds for all regular polygons? And more generally, is it known if it hold for all regular polyhedra and regular polytopes?

I ask "is it known" because a lot of problems in this topic tend to be open.

The Lebesgue differentiation theorem holds for regular families of sets. A regular family is a family of sets that contain sets of arbitrarily small diameters and have the property that there is a constant $$C\geq 1$$ such that for each $$E$$ set in the family there is a ball $$B(x,r)$$ such that $$E\subset B(x,r), \quad |E|>Cr^n$$ (for more details, see e.g. Theorem 2.28 in [1]). All regular polygons form such a family.
Consider the family of all rectangles with sides parallel to coordinate directions. This family is not regular and the Lebesgue theorem fails if $$f\in L^1$$. In fact you can find $$f$$ that its integral is nowhere differentiable for this family. This is a well known result of Saks, see Theorem 2.30 in [1]. However, if $$f\in L^p$$, $$p>1$$, then the integral of $$f$$ is differentiable a.e. with respect to this family. This is a theorem of Zygmund, see Theorem 2.29 in [1].