Defining? No... you asked me to prove that the conditions I gave imply that V is finite-dimensional. So, I did. My proof shows that there exists a finite-dimensional subspace of V that contains V. It uses the fact that any vector space is the union of its finite-dimensional subspaces.

Here's one way: the condition says that the functor V \otimes - is left adjoint to the functor W \otimes -. But it's also left adjoint to hom(V, -), hence we have a canonical iso W \otimes - \cong hom(V, -). In particular, hom(V, -) preserves all colimits. Now V is the union, i.e., the filtered colimit of the system of inclusions V_i \to V_j of its finite-dimensional subspaces. So hom(V, V) is the filtered colimit (union) of subspaces hom(V, V_i). In particular, the identity map 1_V must factor through one of the V_i, hence is finite-dimensional.

Yes, there a number of ways one might think of characterizing finite-dimensionality (including being isomorphic to its double dual!), Noetherian/Artinian hypotheses, etc. But some of these characterizations don't port so well to modules over other commutative rings. The present characterization is equivalent to being finitely generated and projective, for any commutative ring.