If a topological space $X$ of $n$ elements has the discrete topology then all subset $U$ is open and even more, it is regular (i.e., $(\overline{U})^\circ = U$). So, the boolean algebra $RO(X)$ of the regular open subsets of $X$ coincides with $\mathcal{P}(X)\simeq \mathbf{2}^n$ (as boolean algebras). Notice that $X$ is not connected.
Is it possible to choose a connected topological space $X$ (with no necessarily $n$ elements) such that $RO(X)$ has $2^n$ elements?
I think the answer is not. I supposed that such $X$ exists and considered the $n$ atoms $U_n$ of $RO(X)$. So my attemp was to prove that $X$ is not connected via these $\{U_n\}$ (which are disjoint between them) but I could't prove it.
Consider a space $X=\{a,b_1,\dots,b_n\}$ with the topology that a set $U\subseteq X$ is open iff either $a\not\in U$ or $U=X$. This space is connected since there are no disjoint nonempty closed sets. If $A\subseteq X$ is nonempty, then $\overline{A}=A\cup\{a\}$, and so it is easy to see that every open set except $\{b_1,\dots,b_n\}$ is regular. So, $RO(X)\cong\mathcal{P}(\{b_1,\dots,b_n\})$, with the isomorphism mapping every regular open set except $X$ to itself and mapping $X$ to $\{b_1,\dots,b_n\}$.