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$\mathbb RP^3$ 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.

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$\mathbb RP^3$ 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 $S^{2n-1} \subset \mathbb C^n$.