In fact if $(K,\lvert-\rvert)$ is a complete valued field and $L$ is a finite extension of $K$ one can always construct a multiplicative extension of $\lvert-\rvert$ to $L$. Then the argument in KConrad's answer will show that it has to coincide with $\lvert N_{L/K}(-)\rvert^{1/[L:K]}$.
This is theorem 26.3 in Warner's Topological fields. Since the argument there uses many results in topological ring theory I will present an abridged version here for the convenience of the reader.
Let us say that a vector space norm $\|-\|$ on $L$ is multiplicative if $\|1\|=1$ and $\|xy\|\le \|x\|\|y\|$. Moreover we say that $\|-\|$ is spectral if it is multiplicative and we have $\|x^n\|=\|x\|^n$ for every $x\in L$ and $n\ge1$.
Lemma Any two spectral norms on $L$ coincide
Proof: This is the usual argument for the uniqueness of the norms: let $\|-\|_1$ and $\|-\|_2$ be two spectral norms. Then, since any two vector space norms on $L$ are equivalent there exist $c,C>0$ such that $$ c \|x\|_1\le \|x\|_2\le C\|x\|_1$$ foer every $x\in L$. Therefore, by plugging in $x^n$ are taking $n$-th roots we have, by spectrality $$ c^{1/n} \|x\|_1\le \|x\|_2\le C^{1/n}\|x\|_1$$ and letting $n\to \infty$ yields the thesis. $\square$
Lemma There exist a (necessarily unique) spectral norm on $L$.
Proof Take $\|-\|$ a multiplicative norm on $L$ (for example the restriction of the $\ell^1$ operatorn norm on $\operatorname{End}_K(L)$). Then for any $x\in L$ the sequence $\|x^n\|^{1/n}$ is $$\lvert x\rvert:=\lim_{n\ge 1}\|x^n\|^{1/n}$$ (the limit exists because of the multiplicativity of $\|-\|$). We need to prove $\lvert x\rvert$ is a spectral norm on $L$. The only tricky part is to show the triangular identity (note that since $L$ is a field, every multiplicative seminorm is automatically a norm, as its kernel is an ideal). Take $x,y\in L$ and fix $\epsilon>0$. Then there is $N>0$ such that for all $n\ge N$ we have $$\|x^n\|^{1/n}\le \lvert x\rvert+\epsilon\textrm{ and }\|y^n\|^{1/n}\le \lvert y\rvert+\epsilon$$ Therefore there is $C>0$ such that for every $n\ge 0$ we have $$\|x^n\|\le C(\lvert x\rvert+\epsilon)^n\textrm{ and }\|y^n\|\le C(\lvert y\rvert+\epsilon)^n$$ Then $$\|(x+y)^n\|=\left\|\sum_{i=0}^n {n\choose i} x^{n-i}y^i\right\|\le C^2\sum_{i=0}^n {n\choose i} (\lvert x\rvert +\epsilon)^{n-i}(\lvert y\rvert+\epsilon)^i=C^2(\lvert x\rvert +\lvert y\rvert +2\epsilon)^n$$ Finally taking $n$-th roots, letting $n\to \infty$ and $\epsilon\to 0$ proves the thesis. $\square$
Proposition Any spectral norm on $L$ is an absolute value (that is $\lvert xy\rvert=\lvert x\rvert \lvert y\rvert$).
Proof Let $\lvert -\rvert$ be a spectral norm on $L$ and fix $y\in L$. We will show that for every $x\in L$ $\lvert xy \rvert =\lvert x\rvert \lvert y\rvert$. If $y=0$, this is clear, so assume $y\neq 0$, so $\lvert y\rvert\neq 0$. Then define $$ \lvert x\rvert_y:=\lim_{n\to \infty} \lvert xy^n\rvert \lvert y\rvert^{-n}$$ It is easy to verify that $\lvert -\rvert_y$ is a spectral norm on $L$, and therefore $\lvert xy\rvert=\lvert xy\rvert_y$ But we have $\lvert xy\rvert_y=\lvert x\rvert_y \lvert y\rvert$. $\square$
Note that the only fact we have used on $K$ is that any two norms on finite dimensional vector spaces on $K$ are equivalent.