A sufficient (additional) condition is that $B$ be compact for some Hausdorff vector topology for $V$. The proof goes as follows.
Letting $\langle x_i\rangle$ be a Cauchy sequence, it is contained is some $nB$, and so it has some cluster point $y$ there. Given $\varepsilon>0$, there is $i_0$ such that $x_i-x_j\in\frac 12\varepsilon B$ for $i,j\ge i_0$ , and it remains to show that $x_{i_0}-y\in\frac 12\varepsilon B$ . Indeed, if this does not hold, we have $y\not\in x_{i_0}-\frac 12\varepsilon B$ . Since $B$ is compact, it is closed in the Hausdorff case, and so is $x_{i_0}-\frac 12\varepsilon B$ as we have a vector topology. By the cluster point property, there must be some $i\ge i_0$ with $x_i\not\in x_{i_0}-\frac 12\varepsilon B$ , which is impossible.
Edit. Actually, I originally had in mind a more general condition, but I couldn't correctly recall it when writing the answer. Namely, a sufficient condition is that there be a Hausdorff topological vector space $E$ and there an absolutely convex compact set $C$ and a linear map $\ell:V\to E$ with $B=\ell^{-1}[C]$ , and such that we have $y\in{\rm rng\ }\ell$ whenever $y\in E$ is such that $(y+\varepsilon C)\cap({\rm rng\ }\ell)\not=\emptyset$ for all $\varepsilon>0$ . The proof is essentially the same as the one given above with $C$ in place of $B$ . This more general condition applies for example in the case where $B$ is the closed unit ball of $C^k([0,1])$ since we can take $E=({\mathbb R}^{[0,1]})^{k+1}$ and $C=([-1,1]^{[0,1]})^{k+1}$ and $\ell$ given by $y\mapsto\langle y,y'y,''\ldots y^{(k)}\rangle$ .