Does it make sense to complete $V$ with respect to this topology? (Should something stronger be used?)
Will $\omega$ still be non-degenerate on the completion? (Meaning: if $\forall a,\omega(a,b)=0$, then $b=0$.) ...Obviously, yes. In fact, even if $\omega$ was originally degenerate on $V$, the process of completion would "quotient out" the degenerate subspace, as any sequence in it converges to $0$ according to $\omega$.
Is $\omega$ necessarily "strongly symplectic", meaning $a\mapsto\omega(a,\cdot)$ is a bijection between $V$ and its topological dual $V^*$? If not, can we make it so by changing the topology on $V$ (thus changing $V^*$)?
In general, I'm wondering what the conditions and implications are for a symplectic space to be complete and self-dual, similar to a Hilbert space (but without a norm).