Put $f(z) = \sum_{n=0}^{\infty} z^{2^n}$. First note that the set of $\theta \in [0,1)$ such that the binary expansion of $\theta$ has arbitrarily large strings of $100$ consecutive zeros is a set of measure $1$. Now take such a $\theta$, and let $N$ be such that $\{2^{N} \theta\} \le 2^{-100}$ (here $\{x\}$ stands for the fractional part of $x$). Consider $f(e^{-1/2^N + 2\pi i\theta})$ and $f(e^{-1/2^{N+50}+2\pi i \theta})$. The difference of these two quantities is in size
$$
|f(e^{-1/2^N+2\pi i\theta}) - f(e^{-1/2^{N+50} +2\pi i\theta})|
$$
and using the triangle inequality appropriately this
is
$$
\ge \Big|\sum_{n=N+1}^{N+50} e^{-2^n/2^{N+50}} e^{2\pi i 2^n \theta} \Big| - \sum_{n=0}^{N} (e^{-2^{n}/2^{N+50}} - e^{-2^n/2^N}) - \sum_{n=N+51}^{\infty} (e^{-2^{n}/2^{N+50}} -e^{-2^n/2^N}). \tag{1}
$$
The third term in (1) is in size at most $$ e^{-2} + e^{-4} + \ldots \le 1. $$ Since $|e^{-x}-e^{-y}| \le |x-y|$ for $x$ and $y$ in $[0,1]$, the second term in (1) is bounded in size by $$ \sum_{n=0}^{N} \Big( \frac{2^n}{2^N} - \frac{2^n}{2^{N+50}}\Big) \le 2. $$ Finally, since $\{2^n \theta\} \le 2^{-50}$ for $N+1\le n \le N+50$, we see easily that the first term in (1) is $$ \ge \sum_{n=N+1}^{N+50} e^{-2^n/2^{N+50}} - \frac{1}{2^{20}} \ge 40. $$ It follows that the difference in the two values of $f$ is at least $30$ in size, so that the radial limit cannot exist for this $\theta$.