Building on [Pietro Majer's answer][1] to you previous question for a change, consider the following:

Let $g: [0,\infty) \to \mathbb{R}$ be the unique continuous function such that $g(0)=0$, $g'(x) = 1$ for $x \in [\frac{1}{2k},\frac{1}{2k+1}]$ and $g'(x) = -1$ for $x \in [\frac{1}{2k},\frac{1}{2k+1}]$ for any $k\in\mathbb{N}$ as well as (arbitrarily) $g'(x) = 1$ for $x > 1$. Its a bit tedious to write it down, but one can do so even explicitly and find that $|g(x)| < 2x^2$ and obviously $|g'(x)| = 1$ almost everywhere.

Now for any closed set $C\subset \mathbb{R}^n$, consider $f(x) := g(\operatorname{dist}(x,C))$. It is known that $|\nabla \operatorname{dist}(x,C)| =1$ almost everywhere in $\mathbb{R}^n\setminus C$. Assuming that none of the nonzero level-sets of the distance function is of positive $n$-dimensional measure¹, thus $|\nabla f| = 1$ almost everywhere in $\mathbb{R}^n\setminus C$ again.

In contrast, for any $x_0 \in C$, we have $|f(x_0)-f(x)| = |f(x)| \leq g(|x_0-x|) \leq 2|x_0-x|^2$ and thus $f'(x_0) = 0$.

So again if $C$ is chosen as some fractal set of arbitrary Hausdorff-dimension with $\mathcal{H}^n(C) =0$, then this shows that the maximal dimension is $n$ itself.


¹I do not think this can actually ever happen for the distance function, but I could not come up with a quick proof on the fly, so I will only note that it certainly can be avoided for any Cantor-type set.

  [1]: https://mathoverflow.net/a/473610/51695