Are there any infinitely dimensional locally compact spaces? Is the Frechet space of all real sequences locally compact?
Is a Hilbert cube, viewed as a topological metric space locally compact?
 A: Infinite-dimensional ("Hausdorff" is part of the definition) topological vector spaces are never locally compact. The sense of "infinite-dimensional" is ambiguous, beyond this, I think.
Nevertheless, such spaces do admit "substantial" (but not open) subsets which are compact. The classic example, a "Hilbert cube" in $\ell^2$, consisting of $(a_1,a_2,\ldots)$ such that $|a_n|\le {1\over n}$ is compact, and, in fact, has the product topology (as in Tychonoff's theorem). But it is not a nbd of $0$.
For that matter, infinite (even uncountable) topological products of compact spaces are compact, by Tychonoff, if one's sense of "infinite-dimensional" goes that far. In fact, the product topology is disturbingly coarse, despite its sensible mapping properties, so I'd not count such a product as being "seriously infinite-dimensional".
The example of adeles is an instance wherein a somewhat finer topology than a product topology is put on a subset of a cartesian product, producing a locally compact sort-of-infinite-dimensional thing: let $X_i$ be a family of locally compact topological spaces, topological groups for simplicity, and suppose we have an open subset $U_i$ of $X_i$ with compact closure. For some purposes, it is reasonable to consider products indexed by finite subset $S$ of the indices $i$, where $Y_S=\prod_{i\in S} X_i \times \prod_{i\not\in S}K_i$, with the product topology. Then the ascending union of the $Y_S$, really a colimit over $S$, has a finer topology than the subset topology from the product. Nevertheless, it is (fairly obviously) locally compact.
A: Infinite dimensional locally compact spaces?  Are you familiar with the Adeles?
