Which p-adic numbers are also algebraic? What is $\mathbb{Q}_p \cap \overline{\mathbb{Q}}$ ?  
For instance, we know that $\mathbb{Q}_p$ contains the $p-1$st roots of unity, so we might say that $\mathbb{Q}(\zeta) \subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$, where $\zeta$ is a primitive $p-1$st root.  
As a more specific example, $x^2 - 6$ has 2 solutions in $\mathbb{Q}_5$, so we could also say that $\mathbb{Q}(\sqrt{6},\sqrt{-1})\subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$.  
Edit: I removed the motivation for this question (which I think stands by itself), as it will be better as a separate question once I think it through a bit better. 
 A: There's a slightly subtle point near here of which some people are not aware: that it is dangerous (perhaps even nonsensical) to compare algebraic numbers under various different completions. So, to talk about $Q_p\cap \bar Q$, you should be talking about a completion of $Q$ containing $Q_p$, not, e.g., a completion of $Q$ lying inside $C$. I don't think this is what is happening here, but some people may find this interesting.
Now, there are lots of isomorphisms floating around, so usually everything turns out just fine, but sometimes not. Here are two examples.
(1) The following fallacious argument that $e$ is transcendental is from a talk by Gouvêa, "Hensel's p-adic Numbers: early history" (originally due to Hensel himself).
The series expansion of $e^p$ converges in $Q_p$, thus $e$ is a solution to the equation $X^p=1+p\epsilon$, where $\epsilon$ is a $p$-adic unit. So $[Q_p(e):Q_p]=p$ (of course you need to argue that the polynomial is irreducible), and so $[Q(e):Q]\ge p$. Since $p$ was arbitrary, $e$ must be transcendental over $Q$. 
The fallacy is that even though the series for $e$ (and $e^p$) converges in $R$ and $Q_p$, the numbers they converge to are not the same. 
(2) The following is from Koblitz's $p$-adic book, page 83 (with an example and some other fallacious arguments).
It is not true that if an infinite sum of rational numbers (a) converges $p$-adically to a rational number for some $p$ and (b) converges in the real topology to a rational number, then the rational numbers the two series converge to are the same!
A: The field $K_p = \mathbb{Q}_p \cap \overline{\mathbb{Q}}$ is a very natural and well-studied one.  I can throw some terminology at you, but I'm not sure exactly what you want to know about it.
1) It is often called the field of "$p$-adic algebraic numbers".  This comes up in model theory: it is a $p$-adically closed field, which is the $p$-adic analogue of a real-closed field.  In particular, it is elementarily equivalent to $\mathbb{Q}_p$.
2) It is the Henselization of $\mathbb{Q}$ with respect to the $p$-adic valuation, or the 
fraction field of the Henselization of the ring $\mathbb{Z}_{(p)}$ -- i.e., $\mathbb{Z}$ localized at the prime ideal $p$.  
The idea is that this field is not complete but is Henselian -- it satisfies the conclusion of Hensel's Lemma.  Alternately and somewhat more gracefully, Henselian valued fields are characterized by the fact that the valuation extends uniquely to any algebraic field extension.  
Roughly speaking, Henselian fields are as good as complete fields for algebraic constructions but are not "big enough" to have the same topological properties.  For instance, note that $K_p$ cannot possibly be complete with respect to the $p$-adic valuation, because it is countably infinite and without isolated points: apply the Baire Category Theorem.  
