A reference to this answer is Hardy's 'Divergent Series'. Although we do not obtain a magnitude of oscillation, the divergence of $\lim_{x\rightarrow 1-} f(x)$ follows from the following method. Here, we do not need numerical values, nor the Tauberian theorems, just a little bit of Complex Analysis.
After solving this problem, I realized that the same method applies here.
Let $f(x)=\sum_{n=0}^{\infty} (-1)^n x^{2^n}$ as given. Let $g(x)=\sum_{n=0}^{\infty} \frac{(\log x)^n}{(2^n+1)n!}$. Then $$ f(x)+f(x^2)=x, \ \ g(x)+g(x^2)=x. $$ Let $\Phi(x)=f(x)-g(x)$. Then we have $$ \Phi(x)=-\Phi(x^2)=\Phi(x^4). $$ If we prove that $\Phi$ is not a constant, then it follows that $\lim_{x\rightarrow 1-} f(x)$ does not exist.
Taking the principal branch of logarithm, we see that $\Phi(z)=f(z)-g(z)$ is analytic on $D=\{z: |z|<1, \ \ z\notin (-1,0]\}$.
Let $z=re^{2\pi i /3}$, and let $r\rightarrow 1-$. Then we have $$ \Im(f(z))\rightarrow \infty, \textrm{ hence }|f(z)|\rightarrow\infty, $$ $$ g(z) \textrm{ is bounded.} $$ Thus, $\Phi(z)$ is not a constant on $D$. Hence, $\Phi(x), \ \ 0<x<1$ is not a constant.