It is easy to justify the use of non-archimedean norms on number fields from an "inside view" as arising from the prime ideals and are therefore clearly useful a priori.
However, it seems to me that the archimedean norms only make sense if one looks at the embedding of number fields inside $\Bbb C$ but there is no reason to consider this embedding before constructing the norm since $\Bbb C$ is defined as the closure under this norm.
One possible justification would be that there is precisely one archimedean norm on $\Bbb Q$ and this would be satisfactory if one could explain why the axioms we use for a norm should be a priori important in the context of number theory.
My question might be a little unclear and I will be happy to provide clarifications. I am basically looking for a reason why distances should matter in number theory. It is easy to see why we care about them in plane geometry and by extension analysis but they are also very useful in number theory and I do not see why they are useful a priori.
Here is one motivation for considering Archimedean valuations on an ''equal'' level with the non-Archimedean ones, from a purely arithmetic viewpoint. When you do this, you get nice results like the following product formula:
$$ \text{for }x\in \mathbb{Q},\qquad \prod_v |x|_v = |x|_\infty\prod_p |x|_p = 1.$$ The first product is taken over all ''places'' $v$ in $\mathbb{Q}$, where place means equivalence class of absolute values (choosing appropriate normalization), and the second product above uses Ostrowski's result that $\mathbb Q$ has a single Archimedean place, denoted $|x|_\infty$, and that all other places are the non-Archimedean ones corresponding to prime integers.
One way to interpret this product formula is that you actually don't need embeddings into $\mathbb C$ to figure out the Archimedean norm on $\mathbb Q$ -- you just have to multiply together all the non-Archimedean norms, and then take the reciprocal. The ''inside'' and ''outside'' views actually mirror each other.
For number fields $K$ of higher degree, the product formula $\prod_{v\in K} |x|_v = 1$ still holds, but since you generally have more than one Archimedean norm (sometimes denoted $v|\infty$) this does not allow you to express a single Archimedean norm in terms of the non-Archimedean ones. However, it is still true that taken together, the product over non-Archimedean norms gives you equivalent information (that is, the multiplicative inverse) to the product over Archimedean norms.
You can think of this situation as closely analogous to the case of rational functions $k(T)$ over some algebraically closed ground field $k$. Assuming $k$ itself has no ''interesting'' norms, just the trivial one, then there are two types of norms on $k(T)$:
the finite valuations $v_a((T-a)^n) = n$ for any $a\in k$
the degree valuation $v_\infty(f(T)) = - \deg(f)$ for any polynomial $f\in k[T]$.
(Here the valuations correspond to norms via $|\cdot| = \exp(-v(\cdot))$.)
Note that the finite valuation $v_a$ has the geometric interpretation of picking out the order of vanishing, as a zero or pole, of some function on the affine line $\mathbb A^1$, at the closed point $\mathfrak m_a = (T-a)$. We chose to denote the degree valuation by $v_\infty$ because there is also a perfectly nice geometric interpretation here: if we consider the projective line $\mathbb P^1$ as containing $\mathbb A^1$ and one additional point $\infty$, then $v_\infty = -\deg$ picks out the order of vanishing of a function on $\mathbb P^1$ at $\infty$, under the identification of $k(T)$ as the rational function field of both $\mathbb P^1$ and $\mathbb A^1$.
Given a rational function $f\in k(T)$, summing the orders of zeros and poles of $f$ over all points in $\mathbb A^1$ could give you any integer (e.g. when $f = T^n$). However, if we take the same sum over all points in the compact curve $\mathbb P^1$, then the sum is always $0$. This corresponds to the product formula for $K = k(T)$: $$ \sum_{v\in k(T)} v(f) = 0 \quad \Leftrightarrow\quad \prod_{v\in k(T)} |f|_v = 1.$$
The ''geometric'' case $K = k(T)$ has one important advantage over the ''arithmetic'' one $K = \mathbb Q$. In $k(T)$, the infinite valuation $v_\infty$ is isomorphic to any of the finite ones $v_a$ in the sense that there is an automorphism of $k(T)$ sending one to the other, namely $T-a \mapsto \frac1{T-a}$. In $\mathbb Q$, however, there is no such automorphism which exchanges $v_\infty$ with any $v_p$, since $v_p$ is a discrete valuation while $v_\infty$ is not. The geometric analogy suggests that by considering the Archimedean (or ''infinite'') place $v_\infty \in \mathbb Q$ in addition to the $p$-adic places, we are somehow compactifying an arithmetic curve. But the resulting compactification is not smooth and homogeneous like $\mathbb P^1$; rather, the arithmetic curve is highly singular at $\infty$.