Weakly differentiable but classically nowhere differentiable

In any Sobolev space where functions are capable of being infinite at a point, they are capable of being nowhere continuous.

Let $f$ be a function in some Sobolev space which is infinite at 0 (for example $f$ might be $|x|^{-1/3}$ in $H^1(\Omega)$ for $\Omega$ some open ball in $\mathbb{R}^3$), let $q_n$ be an enumeration of the rational points of $\Omega$, and let $f_h(x) = f(x-h)$. Then we can define $$g = \sum_n 2^{-n} f_{q_n}.$$ This $g$ is unbounded on any open set, but its norm is no more than double the norm of $f$, so in particular it is still weakly differentiable.

One can easily verify that the sequence of partial sums converges, and that the result is nowhere continuous, [and that you can't make it continuous by modifying on a set of measure zero].

This is rather a comment on the one-dimensional case and on $W^{1,\infty}_\mathrm{loc}(\Omega)$ than an answer.

In one dimension, this cannot happen. Take a function $f \in L^1_{\mathrm{loc}}(0,1)$, with weak derivative $f' \in L^1_{\mathrm{loc}}(0,1)$. Then, for any $[a,b] \subset (0,1)$, you have $f' \in L^1([a,b])$. Hence, $f$ is absolutely continuous on this interval and hence almost everywhere differentiable.

On the other hand, if $f \in W^{1,\infty}_{\mathrm{loc}}(\Omega)$ for some $\Omega \subset \mathbb{R}^n$, then $f$ is Lipschitz on compact subsets of $\Omega$ and therefore almost everywhere differentiable by Rademacher's theorem.

This shows, that one has to look for counterexamples in at least two dimensions, where the weak derivative is not in $L^\infty_\text{loc}(\Omega)$ (i.e. "nowhere bounded").