In fact, more is true: for any natural $n$ and and any degree $n$ field extension $L$ of any field $K$ and any absolute value $|\cdot|$ on $K$, $|N_K^L(\cdot)|^{1/n}$ is an absolute value on $L$ which extends $|\cdot|$.
Let me give some preliminaries. We regard here a field $L$ and multiplicative function $|\cdot|:L\to [0,\infty)$, that is a function satisfying $|0|=0$, $|1|=1$ and $|xy|=|x||y|$ for every $x,y\in L$. For $C\geq 1$ we say that $|\cdot |$ is a $C$-absolute value if for every $x,y\in L$, $|x+y|\leq C|x+y|$ and if $C=1$ we simply say that $|\cdot|$ is an absolute value. The following is well known.
Lemma 1: A $2$-absolute value is an absolute value.
We say that $|\cdot |$ is a $C$-ultra absolute value if for every $x,y\in L$, $|x+y|\leq C\max\{|x|,|y|\}$ and if $C=1$ we say that $|\cdot|$ is an ultra absolute value.
Note that if $|\cdot |$ is a $C$-ultra absolute value then for every $\alpha>0$, $|\cdot |^\alpha$ is a $C^\alpha$-ultra absolute value hence also a $C^\alpha$-absolute value, while a $C$-absolute value is a $2C$-ultra absolute value. In fact, we have:
Lemma 2: an absolute value $|\cdot|$ is a $|2|$-ultra absolute value.
It follows that every $C$-ultra absolute value is a $|2|$-ultra absolute value, by considering $|\cdot |^\alpha$ for $\alpha=\log_C 2$ and using lemma 1. We get the following.
Corollary: A $C$-ultra absolute value $|\cdot|$ is an absolute value iff $|2|\leq 2$.
Back to the original question, by the discussion above, it enough to show that $|N_K^L(\cdot)|^{1/n}$ is a $C$-absolute value for some $C\geq 1$. This is an exercise (use the fact that the determinant map is bounded on bounded sets of operators).