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 $\alpha\in K\setminus k$ 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 $\mathbb A_k^1(k)$.
Edit: in hindsight I admit the construction of a neighbourhood is not that straightforward, so here are the details. Let $\alpha_1=\alpha,\alpha_2,\dots,\alpha_n$ be the conjugates of $\alpha$ and let $f(x)=(x-\alpha_1)\dots(x-\alpha_n)$. By the argument above there is some $\varepsilon>0$ such that $||x-\alpha_i||\geq\varepsilon$ for all $x\in k$, where $||\cdot||$ is the norm on the algebraic closure of $k$, and hence $||f(x)||\geq\varepsilon^n$. Now we can consider the open subset $U$ of the Berkovich space $\mathbb A_k^1$ consisting of seminorms $|\cdot|$ such that $|f|<\varepsilon^n$. By the above, no point in $\mathbb A_k^1(k)$ lies in $U$.