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. Using your formula for $\mathbb{H}P^{2n+1}$ has signature 1 p_1$, $< p_1^{2n+1}, [\mathbb{H}P^{2n+1}]> = (or -1, depending on your choice of orientation!) 2n-2)^{2n+1} < u,[\mathbb{H}P^{2n+1}]> \neq 0$ if $n>1$, so it $\mathbb{H}P^{2n+1}$ 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). \mathbb{C}P^{2n+1}$. 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!!!
Note: this post has been edited because the original was very false. I claimed that $\sigma(\mathbb{H}P^{2n+1})=1$ which is silly because the middle cohomology is $H^{4n+2}(\mathbb{H}P^{2n+1}) = 0$. Also the signature being odd would have contradicted the fact that $\chi (\mathbb{H}P^{2n+1})$ is even, which is stated in the question.
