If $k$ is not algebraically closed, then $\mathbb A^n_k(k)$ is not doing to be dense with $\mathbb A_k^n$. I'll show this for $n=1$ for simplicity. Take any point $P$ in $\mathbb A^1_k$ with a residue field $K$ which is a proper finite extension $k$. Since $k$ is complete, $k$ is closed as a subspace of $K$, and hence the element $x\in K\setminus k$ corresponding to $P$ (unique up to Galois conjugates) will have an open neighbourhood which doesn't intersect $P$. This neighbourhood gives rise to a nonempty neighbourhood of $P$ in $\mathbb A_k^1$ which does not intersect $\mathbb A_k^1(k)$.