I haven't found any sources that state this explicitly for arbitrary topological vector spaces (most sources are concerned with Frechet spaces, where even weakly holomorphic implies continuous). I'm not sure if this is because it's trivial or because the answer is no. Anyway, I think the answer is yes based on this proof, and was hoping someone could confirm:
Let $\Omega \subseteq \mathbb{C}$ be an open set, $X$ an arbitrary topological vector space, and $f:\Omega \to X$ a strongly holomorphic function, which means that $lim_{w\to z}\frac{f(w)-f(z)}{w-z}$ exists for every $z\in \Omega$.
Given a neighborhood of the origin $U\subseteq X$, choose a balanced neighborhood $V$ such that $V+V \subseteq U$.
Choose $\delta < 1$ small enough that these two conditions both hold for fixed $z$ $$|w-z| < \delta \implies \frac{f(w)-f(z)}{w-z} \subseteq f'(z)+V$$ $$\delta f'(z) \in V$$
Then for $|w-z|<\delta$ we have
$$f(w)-f(z) \subseteq (w-z)(f'(z)+V) \subseteq (w-z)f'(z) + (w-z)V \subseteq V + V \subseteq U$$
So $f$ is continuous.