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 in 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 $y\mapsto\langle y,y',y'',\ldots y^{(k)}\rangle$ .
II Edit. The sufficient condition I gave above is of "extrinsic nature", and as such probably not in the spirit requested in the original question. An "intrinsic" condition, which is (probably) "simple", and in the line already suggested above in the first answer and in the comments, is that for any sequence $\langle x_i:i\in\mathbb N_0\rangle$ in $V$ satisfying $\lbrace 2^{i+2}(x_i-x_{i+1}):i\in\mathbb N_0\rbrace\subseteq B$ , there be $x\in V$ with $\lbrace 2^i(x_i-x):i\in\mathbb N_0\rbrace\subseteq B$ .
However, obviously this is not very practical to be verified in concrete situations. Basing on my experience and intuition, I would generally say that "extrinsic" conditions probably are more convenient than "intrinsic" ones. So, I think the question is good, but the restriction put there on the direction for searching for the answer is wrong. In practice, when one constructs (prospective new) Banach spaces, there is often some surrounding "larger" topological vector space where the new spaces will be continuously injected. In view of this, it is natural to look for extrinsic conditions.
Best Answer
I think the following meets your setup. Let $\mathbb K = \mathbb Z_2$ with the "absolute value" $|0|=0, |1|=1$ (this is non-Archimedean). See http://en.wikipedia.org/wiki/Absolute_value_%28algebra%29
Set $V = \mathbb Z_2^I$ for some index set $I$, with the trivial norm $\|0\|=0$ and $\|x\|=1$ for all other vectors $x$. This satisfies the usual rules, with $\|kx\| = |k|\|x\|$ for $k\in\mathbb K, x\in V$. Clearly $V$ actually have the discrete metric, and so is complete. Now, the closed unit ball is all of $V$, and not compact if $I$ is infinite.
BUT, I could instead define $\|x\|=2$ for $x\not=0$. Still we have a norm. Now the closed unit ball is $\{0\}$; and so is compact. All non-trivial linear maps have norm $1$.
To me, this seems like a very, very silly example, which perhaps shows that the original question needs tweaking with a bit...
If you start with an Archimedean absolute value, then really you have a subfield of $\mathbb C$ or $\mathbb R$ (which must contain $\mathbb Q$). By continuity, I then think you can turn $V$ into a $\mathbb R$ vector space, with a "norm" in the usual sense. Then if $V$ has a compact unit ball, it must be finite dimensional (over $\mathbb R$). So your example of $\mathbb R^2$ over $\mathbb Q$ is in a sense all that can happen.