By an absolute value over $\mathbb{C}$, I mean a function $\mathbb{C}\to \mathbb{R}_{\ge 0}$ such that:
$\bullet$ $|x| = 0\Longleftrightarrow x=0$;
$\bullet$ for all $x,y\in\mathbb{C}$, we have $|xy| = |x||y|$;
$\bullet$ for all $x,y\in\mathbb{C}$, we have $|x+y|\le |x|+|y|$.
Two absolute values $|\cdot|_1$ and $|\cdot|_2$ are called equivalent if there exists a constant $c\in\mathbb{R}^+$ such that $|x|_2 = |x|^c_1$ for all $x\in\mathbb{C}$.
Suppose that one finds to characterize all nontrivial absolute values over $\mathbb{C}$ up to equivalence, just like Ostrowski's theorem does for $\mathbb{Q}$.
If $|\cdot|$ is an archimedean absolute value over $\mathbb{C}$, by another Ostrowski's theorem there exists an embedding of valued fields (that is, embedding of fields and topological spaces) $\sigma:(\mathbb{C},|\cdot|)\to(\mathbb{C},|\cdot|_0)$, where $|\cdot|_0$ is the usual absolute value over $\mathbb{C}$. WLOG we can suppose that $\sigma:(\mathbb{C},|\cdot|)\to(\sigma(\mathbb{C}),|\cdot|_0)$ is a bijection and an isometry, since $|\cdot|$ and $|\sigma(\cdot)|_0$ are equal up to a constant exponent.
Now suppose that $|\cdot|$ is nonarchimedean. By Ostrowski's theorem, the restriction of $|\cdot|$ over $\mathbb{Q}$ is equivalent to the $p$-adic absolute value for some prime $p$. Must there exist an embedding of valued fields $\sigma:(\mathbb{C},|\cdot|)\to\mathbb{C}_p$, where $\mathbb{C}_p$ is the canonical completion of $\overline{\mathbb{Q}}_p$ (that is, consisting of the equivalent classes of Cauchy sequences over $\overline{\mathbb{Q}}_p$)? In other words, can we characterize all absolute values over $\mathbb{C}$ (or equivalently, over $\mathbb{C}_p$) whose restriction over $\mathbb{Q}$ is the $p$-adic absolute value?
Note 1: If the general case is too far to be determined, we can assume that $|\cdot|$ is complete. It should be noted that, suppose that an embedding $\sigma:(\mathbb{C},|\cdot|)\to\mathbb{C}_p$ exists, $|\cdot|$ is complete if and only if $\sigma$ is actually an isomorphism: "$\Leftarrow$" is obvious. "$\Rightarrow$": $\operatorname{Im}\sigma$ is a (topologically) closed subfield of $\mathbb{C}_p$ that is algebraically closed (being isomorphic to $\mathbb{C}$ as fields). Since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, we have $\mathbb{Q}_p\subset\operatorname{Im}\sigma$, and then $\overline{\mathbb{Q}}_p\subset\operatorname{Im}\sigma$, then $\operatorname{Im}\sigma=\mathbb{C}_p$ since $\overline{\mathbb{Q}}_p$ is dense in $\mathbb{C}_p$.
Note 2: If the desired result is true, then an interest result would be that there is no nontrivial complete nonarchimedean absolute value over $\mathbb{R}$: suppose that $|\cdot|$ is one. It is standard that a complete absolute value extends uniquely to any algebraic extension. Write the extension of $|\cdot|$ over $\mathbb{C}$ as $|\cdot|'$. There exist an embedding of valued fields $\sigma:(\mathbb{C},|\cdot|')\to\mathbb{C}_p$. We have that $\sigma(\mathbb{R})$ is a (topologically) closed subfield of $\mathbb{C}_p$ that is isomorphic to $\mathbb{R}$ as fields. But we would then have $\mathbb{Q}_p\subset\sigma(\mathbb{R})$, which is impossible because $\mathbb{R}$ does not contain a field isomorphic to $\mathbb{Q}_p$ for some prime $p$ (negative numbers can have square roots in $\mathbb{Q}_p$ for every $p$ but never in $\mathbb{R}$).
Best Answer
The answer to your question "Must there exist..." is no; things are not that simple.
Take $\mathbb{C}_p$ and adjoin an indeterminate $\theta$. The absolute value $|\cdot|$ on $\mathbb{C}_p$ can be extended to $\mathbb{C}_p(\theta)$ by setting $|\sum_{i\ge 0} a_i \theta^i|:=\max_i |a_i|$ and $|f(\theta)/g(\theta)|:=|f(\theta)|/|g(\theta)|$ $(g\ne 0)$. Take the completion of this field with respect to $|\cdot|$; then take the algebraic closure, extending $|\cdot|$ in the only possible way, and call the resulting field $F$. (You could instead let $F$ be the completion of this field with respect to this absolute value, if you prefer.)
Suppose that we set $G:=\mathbb{C}_p$ and assume there is a continuous field embedding $\sigma:F\rightarrow G$. Then $\sigma$ must be the identity on $\mathbb{Q}$ and, also, by continuity, on $\mathbb{Q}_p$. Also, on any finite normal extension $E$ of $\mathbb{Q}_p$ in $F$, $\sigma$ must send $E$ onto the copy of $E$ in $G$, although not necessarily as the identity; but since $|\cdot|$ can be defined on $E$ by $|x|=|N_{E/\mathbb{Q}_p}(x)|^{1/[E:\mathbb{Q}_p]}$, we do have $|\sigma(x)|=|x|$ so $\sigma$ is an isometry on $E$. Hence $\sigma$ is an isometric field isomorphism on $\overline{\mathbb{Q}_p}$. It follows by continuity that $\sigma$ extends to an isometric isomorphism from $\mathbb{C}_p$ to itself, leaving nowhere to map $\theta$. Since $\mathbb{C}$ and $F$ are isomorphic as abstract fields, as both are algebraic closures of the extension of $\mathbb{Q}$ by $2^{\aleph_0}$ independent transcendentals, this gives a negative answer to your first question.