The answer should be: if $n = m$, then the image is the roots of unity of order $p^{n-1}$. If $n = m - 1$ then the norm map should be surjective on the roots of unity.

In the first case, since all roots of unity $\zeta$ are already contained in $k$ and $[L:K] = p$, then $N_{L/K}(\zeta) = \zeta^p$, and $\zeta \mapsto \zeta^p$ is surjective onto the index-$p$ subgroup of $p^{n-1}$-order roots of unity.

In the second case, fix $\zeta \in K$, and there is a $\zeta_0$ in $L$ with $\zeta_0^p = \zeta$. Then the minimal polynomial of $\zeta_0$ is $x^p - \zeta = 0$, so the norm of $\zeta_0$ is $\zeta$ (unless $p = 2$, in which case it is $-\zeta$, but we still get surjectivity).

The answer should be: if $n = m$, then the image is the roots of unity of order $p^{n-1}$. If $n = m - 1$ then the norm map should be surjective on the roots of unity.

In the first case, since all roots of unity $\zeta$ are already contained in $K$ and $[L:K] = p$, then $N_{L/K}(\zeta) = \zeta^p$, and $\zeta \mapsto \zeta^p$ is surjective onto the index-$p$ subgroup of $p^{n-1}$-order roots of unity.

In the second case, fix $\zeta \in K$, and there is a $\zeta_0$ in $L$ with $\zeta_0^p = \zeta$. Then the minimal polynomial of $\zeta_0$ is $x^p - \zeta = 0$, so the norm of $\zeta_0$ is $\zeta$ (unless $p = 2$, in which case it is $-\zeta$, but we still get surjectivity).