My intuition for this comes from the fact that a plane would be its own boundary, so therefore $\mathbb R^n$ should be it's own boundary.
The context of my question is thinking about "divergence free vector fields tangent to the boundary of $M$" for some manifold $M$ and trying to make a concrete example in $\mathbb R^n$.
Alternatively you can also see this by observing that a neighborhood of any point in a set's boundary must intersect its complement, which in this case is empty. So the boundary must be empty.
The boundary of $B=\bar B-O(B)$ where $\bar B$ is the adherence of $B$ and $O(B)$ is the interior of $B$ is the adherene of $B$, $\mathbb{R}^n$ is closed and open in $\mathbb{R}^n$, so we deduce that its boundary is empty.
