Edit: The statement below is wrong. The signature is 0. However the Pontryagin number argument still holds when $n>1$ (obviously the four sphere bounds a ball which is orientable). I will rewrite the post tomorrow.
Jason, this not an answer, just an observation. $\mathbb{H}P^{2n+1}$ has signature 1 (or -1, depending on your choice of orientation!) so it cannot be the boundary of an oriented manifold, unlike the examples you give for $\mathbb{R}P^{2n+1}$ and $\mathbb{C}P^{2n+1}$ (alternatively, one of the Pontryagin numbers, which can be computed from the classes in your comment is not zero). The point is that filling spherical fibres in oriented bundles will not work.
By the way, this is my first post in Math Overflow. Yay!!!