In fact, more is true: for any local field $K$, any degree $n$ field extension $L$ of $K$ and any absolute value $|\cdot|$ on $K$, $|N_K^L(\cdot)|^{1/n}$ is the unique absolute value on $L$
which extends $|\cdot|$. In particular, it is a norm on $L$.
In this post I intend to give a proof of this fact which
does not rely on properties of $K$ other than local compactness.
The proofs of the technical Lemma 1 and Lemma 2 are postponed to the end, not to interrupt the reading flow.
Let me give some preliminaries.
We regard here a field $F$ and a multiplicative function $|\cdot|:F\to [0,\infty)$,
that is a function satisfying $|0|=0$, $|1|=1$ and $|xy|=|x||y|$ for every $x,y\in F$.
For $C\geq 1$ we say that $|\cdot |$ is a
$C$-absolute value if for every $x,y\in F$, $|x+y|\leq C(|x|+|y|)$
and if $C=1$ we simply say that $|\cdot|$ is an absolute value.
The following is well known.
It is an easy exercise to check that if $|\cdot|$ is a $C$-absolute value and $\alpha\in(0,1]$ then $|\cdot |^\alpha$ is a $C^\alpha$-absolute value. However, this does not work in general for $\alpha>1$. To remify this, we study a more homogenous condition.
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.
Now we indeed have that if $|\cdot |$ is a $C$-ultra absolute value then for every $\alpha>0$,
$|\cdot |^\alpha$ is a $C^\alpha$-ultra absolute value.
The two definitions relate trivially: a $C$-ultra absolute value is a $C$- absolute value while a $C$-absolute value is a $2C$-ultra absolute value.
In particular, every absolute value is a $2$-ultra absolute value.
The following, however, is less trivial.
Proof: Set $\alpha=\log_C 2$ and consider the $2$-ultra absolute value $|\cdot |^\alpha$.
It is trivially a $2$-absolute value, thus an actual absolute value by Lemma 1. By Lemma 2 it is a $\max\{1,|2|^\alpha\}$-ultra absolute value.
Taking now the $1/\alpha$-power, we get that $|\cdot|$ is indeed a $\max\{1,|2|\}$-ultra absolute value.
$\square$
Proof: If $|\cdot|$ is an absolute value then clearly $|2|=|1+1|\leq |1|+|1|=2$. Assume $|\cdot|$ is a $C$-absolute value and $|2|\leq 2$.
Then $|\cdot|$ is a $2C$-ultra absolute value, thus by Corollary A, it is a $\max\{1,|2|\}$-ultra absolute value,
hence a $2$-ultra absolute value, as $\max\{1,|2|\}\leq 2$. In particular,
$|\cdot|$ is a $2$-absolute value, thus it is an actual absolute value by Lemma 1.
$\square$
Proof: This follows from Corollary B, as 2 belongs to the subfield.
$\square$
Proof: This follows easily from the continuity of $N$ at 0 and the fact that $N(x^{-1})=N(x)^{-1}$.
$\square$
Proof:
The unit ball $B\subset K$ is compact, hence so is $N^{-1}(B)\subset L$
and its shift $1+N^{-1}(B)$. It follows that the image in $[0,\infty)$ under $|N(\cdot)|$ of $1+N^{-1}(B)$ is bounded by some $C$, thus for $z\in L$,
$$ |N(z)|\leq 1 \Rightarrow |N(z)+1|\leq C. $$
It follows that $|N(\cdot)|$ is a $C$-ultra absolute value.
Indeed, for $x,y\in L$, assuming wlog $|x|\leq |y|$ and setting $z=xy^{-1}$ we have
$$ |N(x+y)|=|N(y)||N(z)+1|\leq C|N(y)||N(z)+1|= C(|N(x)|+|N(y)|).$$
It follows that $|N(\cdot)|^{1/n}$ is a $C^{1/n}$-ultra absolute value,
thus by Corollary C, it is an actual absolute value.
$\square$
I will now provide the proofs of Lemma 1 and Lemma 2.
Proof of Lemma 1:
Assume $|\cdot|$ is a 2-absolute value.
We first observe that for every natural $j$, and every $2^j$ elements $x_1,\ldots, x_{2^j}\in F$, we have
$$ |\sum_{i=1}^{2^j} x_i|\leq 2^j \sum_{i=1}^{2^j} |x_i|.$$
Indeed, this follows easily by induction on $j$.
Picking any natural $m$ and considering $j$ such that $2^{j-1}<m\leq 2^j$,
we get that for every $m$ elements $x_1,\ldots, x_m\in F$, we have
$$ |\sum_{i=1}^{m} x_i|\leq 2m \sum_{i=1}^{m} |x_i|.$$
Indeed, this follows by adding $x_i=0$ for $m<i\leq 2^j$ to the list
and observing that $2^j\leq 2m$.
In particular, by taking $x_i=1$, we now have for every natural $m$, $|m|\leq 2m$.
We now pick arbitrary $x,y\in F$ and a natural $n$ and make the following estimates:
$$ |x+y|^n=|(x+y)^n|=\left|\sum_{k=0}^n {n\choose k} x^ky^{n-k}\right|
\leq 2(n+1) \sum_{k=0}^n \left|{n\choose k}\right| |x^k||y^{n-k}| \leq
$$
$$ 4(n+1) \sum_{k=0}^n {n\choose k}|x^k||y^{n-k}|=4(n+1)(|x|+|y|)^n.$$
Taking $n$th root an letting $n\to \infty$, we get indeed,
$$ |x+y|\leq |x|+|y|.$$
$\square$
For the proof of Lemma 2 we will need the following.
Claim: Assume $|\cdot|$ is an absolute value.
Then for every pair of naturals $k< m$, we have
$|k|\leq \max\{1,|m|\}$.
Proof of the Claim:
Fix a natural $n$
and expand $k^n$ on base $m$,
$$ k^n=\sum_{i=0}^{n-1} a_i m^i $$
for some integers $0\leq a_i<m$. Note that indeed it is enough to consider indexes bounded by $n-1$, as $k<m$.
Thus we have
$$ |k|^n=|k^n|=|\sum_{i=0}^{n-1} a_i m^i|\leq
\sum_{i=0}^{n-1} a_i |m|^i < m \sum_{i=0}^{n-1} |m|^i. $$
If $|m|\leq 1$ then we get $ k^n < mn $
and taking $n$th root and $n\to \infty$ we conclude that indeed $|k|\leq 1\leq \max\{1,|m|\}$.
If $|m|> 1$ then we get
$$ |k|^n <m \sum_{i=0}^{n-1} |m|^i = \frac{m(|m|^n-1)}{|m|-1}$$
and taking $n$th root and $n\to \infty$ we conclude that indeed $|k|\leq |m|\leq \max\{1,|m|\}$.
This proves the claim.
$\square$
Proof of Lemma 2:
We assume $|\cdot|$ is an absolute value.
We pick arbitrary $x,y\in F$ and a natural $n$.
We note that for $0\leq k\leq n$ we have $|x^k||y^{n-k}|\leq \max\{|x|,|y|\}^n$. Also we have ${n\choose k}\leq 2^n$,
thus by the claim $|{n\choose k}|\leq \max\{1,|2^n|\}=\max\{1,|2|\}^n$.
We get the following estimates:
$$ |x+y|^n=|(x+y)^n|=\left|\sum_{k=0}^n {n\choose k} x^ky^{n-k}\right|
\leq \sum_{k=0}^n \left|{n\choose k}\right| |x^k||y^{n-k}| \leq
$$
$$ \sum_{k=0}^n \max\{1,|2|\}^n\max\{|x|,|y|\}^n = (n+1)\max\{1,|2|\}^n\max\{|x|,|y|\}^n.$$
Taking $n$th root an letting $n\to \infty$, we get indeed,
$$ |x+y|\leq \max\{1,|2|\}\max\{|x|,|y|\}.$$
$\square$
