Here,$\{q_i\}$ an enumeration of the rationals in the closed unit interval.Take $y=q_j\in \mathbb{Q}\cap [0,1]$.
I tried to gleen something from the following manipulation
\begin{align*} &\ \ \ \lim_{x\rightarrow q_j}\frac{1}{x-q_j}\sum_i 2^{-i}(|x-q_i|-|q_j-q_i|)\\ &=\lim_{x\rightarrow q_j}\frac{1}{x-q_j}(2^{-j}|x-q_{j}|+\sum_{i\ne j} 2^{-i}(|x-q_i|-|q_j-q_i|))\\ &=2^{-j}+\lim_{x\rightarrow q_j}\frac{1}{x-q_j}\sum_{i\ne j} 2^{-i}(|x-q_i|-|q_j-q_i|) \end{align*} Which I think will blow up, since the sum is some constant and the denominator blows up, with no chance of the differences in the numerator vanishing.
Is my work correct? How can I show this a little more rigorously than just saying some words? For example, am I justified in commuting the limit and the sum?
You can not bring $|x-q_j|$ out of the limit. In fact the non-differentability comes from the different left and right derivatives at $q_j$, not from the blowup, which does not occur: $f$ is an absolutely convergent series of convex functions, therefore it is convex. In particular the left and right derivatives exist at every point.
If you fix a rational $q_j$, then the difference between the left and right derivative of $f$ at $q_j$ is at least $2^{-j+1}$, coming from $2^{-j}|x-q_j|$: adding other convex functions can just increase the difference (use monotonicity of incremental ratios).
We can see that indeed the differences in the numerator do cancel out: when $x$ approaches $q_j$, for more and more rationals $q_i$ it will be true that $x$ and $q_j$ are on the same side of $q_i$, therefore yielding $\big||x-q_i|-|q_j-q_i|\big|=|x-q_j|$; while if they are not on the same side, say $q_j<q_i<x$, then $|x-q_i|-|q_j-q_i|\leq 2|x-q_j|$.