An answer to the question in the second paragraph (what kind of objects should a quantum field theory assign to codimension 2 manifolds?) can be found in my paper http://arxiv.org/abs/1304.7328v2.
I will argue that this also provides an answer to the OP's main question (what is a possibly infinite dimensional 2-Hilbert space?)
The answer is: it's a von Neumann algebra (or, possibly better, the module category of a von Neumann algebra).
Note that the kind of von Neumann algebras I have in mind are type $III$ factors and that for such an algebra, there is very little difference between the algebra and its representation category. If one excludes the zero module (and modules on non-separable Hilbert spaces), then the von Neumann algebra itself, viewed as a one object category, is equivalent (in the usual sense of equivalence of categories) to its representation category. In other words, such a von Neumann algebra has only one non-zero module up to isomorphism (excluding non-separable modules), and the endomorphism algebra of this unique module is the von Neumann algebra you started with. Note that this unique module is never irreducible: if you direct sum it with itself, you get again itself.
One can actually be more precise: the von Neumann algebras that show up in quantum field theory are always hyperfinite type $III_1$ factors (unless they are Morita equivalent to something finite dimensional). By a famous theorem of Connes and Haagerup, there is only one hyperfinite type $III_1$ factor (subject to appropriate separability assumptions) up to isomorphism. Thus, if one takes the point of view that an ``infinite dimensional 2-Hilbert space'' is the same thing as a hyperfinite type $III_1$ factor, then the Connes-Haagerup theorem says that there's a unique infinite dimensional 2-Hilbert space, up to isomorphism. This is the analog of the well known and easy fact according to which there exists only one separable infinite dimensional Hilbert space, up to isomorphism.
Now one might ask: what is the 2-inner product on $M$-Mod? (here, $M$ is any von Neumann algebra).
For that, note that there is a complex-antilinear functor $M$-Mod $\to$ $M^{op}\!$-Mod given by sending a Hilbert space $H$ with action of $M$ to its complex conjugate $H\mapsto \overline H$. Here, the right action of $m\in M$ on $\overline H$ is given by $\overline\xi m:=\overline{m^*\xi}$. Given $H\in M$-Mod and $K\in M^{op}\!$-Mod, one can form the relative tensor product $K\boxtimes_M H$ (also explained here), which is some kind of fancy version of the tensor product over a ring. That's the 2-inner product. Namely, for two $M$-modules $H_1$ and $H_2$, we define $$\langle H_1,H_2\rangle := \overline{H_1}\boxtimes_MH_2 \in \text{Hilb}.$$
To contrast with Turion's answer, let me emphasise that a 2-Hilbert space is not enriched in Hilbert spaces. It is enriched in von Neumann algebras.
Also, a 2-Hilbert space doesn't have a product with $\text{Vect}$ for scalar multiplication. It has a product with $\text{Hilb}$.
As a final remark, let me point out that my ``2-Hilbert spaces'' are suitable for representing 2-Groups (topological groups with a $U(1)$-valued 3-cocycle in the sense of Segal), just like Hilbert spaces are good for representing centrally extended groups (topological groups with a $U(1)$-valued 2-cocycle).