Jason, this not an answer, just an observation. Using your formula for $p_1$, $< p_1^{2n+1}, [\mathbb{H}P^{2n+1}]> = (2n-2)^{2n+1} < u,[\mathbb{H}P^{2n+1}]> \neq 0$ if $n>1$, so $\mathbb{H}P^{2n+1}$ cannot be the boundary of an oriented manifold, unlike the examples you give for $\mathbb{R}P^{2n+1}$ and $\mathbb{C}P^{2n+1}$. 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.
I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and upvoters. I thought it more helpful to correct it than delete outright, but read critically.
If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f_0$ and $f_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{Z}/(p^l)$ for all primes $p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. To see this, write $H^n(Y;\mathbb{Z})$ as a direct sum of $\mathbb{Z}^{r}$ and various primary summands $\mathbb{Z}/(p^k)$, and note that the summand $\mathbb{Z}/(p^k)$ restricts injectively to the mod $p^l$ cohomology when $l\geq k$. One can take only those $p^l$ such that there is $p^l$-torsion in $H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is pretty implausible, and is indeed wrong.)
We can try to apply this to $Y=BG$, for $G$ a compact Lie group. For example, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology), and so rational characteristic classes suffice. In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, because, as Allen Hatcher has pointed out
in this recent answer,
all the torsion is actually 2-torsion.
It's sometimes worthwhile to consider the integral Stiefel-Whitney classes $W_{i+1}=\beta_2(w_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W_3=0$.
[I'm sceptical of your example in $2\mathbb{CP}^2$. So far as I can see, $3a+3b$ squares to 18, not 6, and indeed, $p_1$ is not a square.]
Best Answer
$\mathbb RP^3$ double-covers the lens space $L_{4,1}$, so it's the boundary of the mapping cylinder of that covering map.
In general $\mathbb RP^n$ for $n$ odd double-covers such a lens space. So in general $\mathbb RP^n$ is the boundary of a pretty standard $I$-bundle over the appropriate lens space. To be specific, define the general $L_{4,1}$ as $S^{2n-1} / \mathbb Z_4$ where $Z_4 \subset S^1$ are the 4-th roots of unity, and we're using the standard action of the unit complex numbers on an odd dimensional sphere $S^{2n-1} \subset \mathbb C^n$.
Edit: generalizing Tim's construction, you have the fiber bundle $S^1 \to S^{2n-1} \to \mathbb CP^{n-1}$. This allows you to think of $S^{2n-1}$ as the boundary of the tautological $D^2$-bundle over $\mathbb CP^{n-1}$. You can mod out the whole bundle by the antipodal map and you get $\mathbb RP^{2n-1}$ as the boundary of the disc bundle over $\mathbb CP^{n-1}$ with Euler class $2$. So this gives you an orientable manifold bounding $\mathbb RP^{2n-1}$ while my previous example was non-orientable.