If 'compactly generated' is used in the loose sense that there is a compact subset that generates a dense subgroup, then all bets are off. However, if there is a generating set in the strict sense that is compact, then we can at least say 'completely metrisable $\Rightarrow$ locally compact':

We may assume that the compact generating set $X$ contains inverses (since the set of inverses of $X$ is also compact), and we also note that $X^n$ is compact for all $n$. Thus the group $G$ is a union of a countable ascending chain of compact sets. One of these has non-empty interior, by the Baire category theorem.

Edit: $(\mathbb{Q},+)$ with the order topology is compactly generated (by the set $\{1/n \mid n \in \mathbb{N}\} \cup \{0\}$) but not locally compact. I would say this is the 'easiest' example that answers the OP's question. (It's also a good example of an infinite totally disconnected group with no proper open subgroups, which is something that cannot occur for locally compact groups.) The groups mentioned by YCor in his comment have similar topological properties.

Old answer: If 'compactly generated' is used in the loose sense that there is a compact subset that generates a dense subgroup, then all bets are off. However, if there is a generating set in the strict sense that is compact, then we can at least say 'completely metrisable $\Rightarrow$ locally compact':

We may assume that the compact generating set $X$ contains inverses (since the set of inverses of $X$ is also compact), and we also note that $X^n$ is compact for all $n$. Thus the group $G$ is a union of a countable ascending chain of compact sets. One of these has non-empty interior, by the Baire category theorem.