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 $|\cdot|$. 

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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 $|2|\leq 2$. 

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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).