The function $f(x)$ is $\mu([0,x])$ where $\mu$ is the radon measure $\sum_{n\in\mathbb{Z} _ +} 2^{-n}\delta _ {q_n}$, and $\mu$ is singular w.r.to the Lebesgue measure $\lambda$ (in fact, $\operatorname{supp}(\mu)=(0,1)\cap\mathbb{Q}$). So the absolutely continuous part $\mu ^ a$ w.r.to $\lambda$ is zero, and the Radon-Nikodym derivative $d\mu ^ a/d \lambda$ is also zero; but this coincides a.e. with the derivative of $f$.
Note that (depending on the particular chosen enumeration of $(0,1)\cap\mathbb{Q}$) there might be infinitely many irrational points $x$ where $f$ is continuous and derivable with any value of $f'(x)$; a Lebesgue null set though.

**edit**. A more elementary argument. Consider the nested family of open nbd's of $(0,1)\cap\mathbb{Q},$

$$A_\epsilon:=\cup_{n\in\mathbb{Z} _ + } (r_n- \epsilon 2^{-n/3},r_n+ \epsilon 2^{-n/3}), \qquad \epsilon > 0.$$
So $|A _\epsilon|=O(\epsilon)$ and $A:=\cap _ {\epsilon > 0} A _ \epsilon$ has measure zero. Let $x \in (0,1) \setminus A$: There exists $\epsilon > 0$ such that for any $n\in\mathbb{Z} _ +$ there holds $ \epsilon 2^{-n/3}\le |x-r _ n|$. Thus, for any $y\in (0,1)$

\begin{align*}
|f(x)-f(y)| &\le \sum_{|x- r _ n|\le|x- y| } 2^{-n}\\
&= \frac{1}{\epsilon^2}\sum_{|x- r _ n|\le|x- y| } 2^{-n/3}(\epsilon 2^{-n/3})^2\\
&\le \frac{1}{\epsilon^2}\bigg(\sum_{n=1}^\infty 2^{-n/3}\bigg)|x-y|^2\\
&= \frac{|x-y|^2}{\epsilon^2(2^{1/3}-1))}
\end{align*} showing that $f'(x)=0.$