Since questions recirculate to the front page forever if left unanswered, I will amalgamate the comments into an answer, which I've made community wiki.
Firstly, by the characterization of the Cantor space as the only metrizable Stone space with no isolated points (up to homeomorphism), $\newcommand{\Z}{\mathbb{Z}}\Z_p$ is homeomorphic to the Cantor space. A proof that the Cantor space is not extremally disconnected can be found here: https://proofwiki.org/wiki/Cantor_Space_is_not_Extremally_Disconnected
In fact, any convergent sequence in a Stonean space $X$ is eventually constant. If we have a sequence that is not eventually constant, we can pass to a subsequence $\newcommand{\N}{\mathbb{N}}(x_i)_{i \in \N}$ such that the mapping $\N \rightarrow X$ defined by $i \mapsto x_i$ is injective. Using the Hausdorffness of $X$, we can build up a disjoint sequence $(U_i)_{i \in \N}$ of open sets such that $x_i \in U_i$ but not in any other. Then $U = \bigcup_{i=0}^\infty U_{2i}$ and $V = \bigcup_{i=0}^\infty U_{2i+1}$ are disjoint open sets, but $x \in \overline{U} \cap \overline{V}$, so they do not have disjoint closures, which contradicts $X$ being Stonean. It follows that any metrizable stonean space is discrete.
To answer the question in your second-to-last paragraph, proving the existence of infinite Stonean spaces requires the axiom of choice, so there are no "explicit" examples. What I mean by this is that ZF + DC (dependent choice) + "all Stonean spaces are finite" is relatively consistent to ZF. So, unless you make a set-theoretic assumption, you can't find a simpler description of an infinite Stonean space than "the Stone space of the complete Boolean algebra $A$" or "the Gelfand spectrum of the commutative AW*-algebra $B$" or "the Yosida spectrum of the Dedekind-complete Riesz space $C$".
To prove the relative consistency of ZF + DC + "all Stonean spaces are finite", let $X$ be an infinite Stonean space, and $A$ be the set of clopens of $X$, as a complete Boolean algebra (the join of a set of clopens is the closure of their union, this requires no AC). If $A$ were finite, then the intersection of all clopens containing a point would be clopen, so $X$ would be discrete, which contradicts it being compact and infinite. So $A$ is an infinite complete Boolean algebra.
Now, under DC, every infinite complete Boolean algebra $A$ contains a countable disjoint sequence, by the following argument. If $A$ contains infinitely many atoms, we use DC to get a sequence $(a_i)_{i \in \N}$ of disjoint atoms of $A$ (apply DC to the poset of finite disjoint sequences of atoms ordered by extension). If it does not, let $b$ be the join of the atoms, and so the complete Boolean algebra $B = \downarrow(\lnot b)$ is atomless and infinite. If we apply DC to the poset of non-zero elements of $B$, ordered by $<$ (not by $\leq$), then we get a strictly descending sequence $(b_i)_{i \in \N}$ of elements of $B$, which are also elements of $A$. So $a_i = b_i \land \lnot\bigvee_{j = i+1}^\infty b_j$ defines a disjoint family $(a_i)_{i \in \N}$ in $A$.
By the previous paragraph, there must be a disjoint family $(C_i)_{i \in \N}$ of clopens in $X$. If $U = \bigcup_{i \in \N}C_i$ were closed, it would be compact, and therefore the union of some finite subfamily, which it can't be by disjointness. So there exists $x \in \overline{U} \setminus U$. Define $u \subseteq \mathcal{P}(\N)$ by $$ u = \left\{ S \subseteq \N \,\left\lvert\, x \not\in \overline{\bigcup_{i \in S}C_i} \right.\right\} $$ It is easy to verify from properties of the closure (monotonicity and preservation of finite unions) that $u$ is an ultrafilter and all finite subsets of $\N$ are in $u$, so it is non-principal.
Now, by constructing a model, Solovay and Pincus proved that ZF + DC (dependent choice) + "there exist no non-principal ultrafilters on any set" is relatively consistent to ZF in Theorem 2 of that article (apparently this was previously proved by Andreas Blass, but I can't find the article anywhere). So by the previous paragraphs, in Solovay and Pincus's model all Stonean spaces are finite.