Here's the solution I expected: Let $G=\operatorname{Gal}(\Bbb{F}_{p^2}/\Bbb{F}_p)$ by construction, the cokernel of the norm is the 0-th Tate cohomology $\hat{H}^0(G,\Bbb{F}_{p^2})$, since $G$ is cyclic, periodicity of the Tate cohomology implies that $\hat{H}^0(G,\Bbb{F}_{p^2}) \cong \hat{H}^2(G,\Bbb{F}_{p^2})\cong H^2(G,\Bbb{F}_{p^2})$ Now the latter group is a relative Brauer group, so it classifies division algebras with center $\Bbb{F}_{p^2}$ and degree $2$.
But by Wedderburn's little theorem, all finite division rings are fields, so the group must be zero, thus the norm is surjective

Write $q=p^n$, $l=p^k$. Choose an algebraic closure of $\Bbb{F}_p$ and embed everything into that. Note that $\Bbb{F}_l$ is a splitting field of $x^l-x$ and consists of all element in $\overline{\Bbb{F}_p}$ such that $x^l=x$. This implies that the subset of elements $x$ in $\Bbb{F}_q$ that satisfy $x^l=x$ is actually $\Bbb{F}_l \cap \Bbb{F}_q= \Bbb{F}_{p^k} \cap \Bbb{F}_{p^n}$.
Now for every $m \in \Bbb N$, there's a unique finite field inside our chosen algebraic closure of degree $m$ over $\Bbb{F}_p$. Furthermore, we have $\Bbb{F}_{p^a} \subset \Bbb{F}_{p^b}$ iff $a \mid b$.