If $(X,\mathcal T)$ is a topological space then $\mathcal F\subseteq\mathcal T$ is called an open filter on $X$ is (i) $X\in\mathcal F$, $\emptyset\notin\mathcal F$; (ii) $A,B\in\mathcal F$ $\Rightarrow$ $A\cap B\in\mathcal F$; $A\in\mathcal F$, $A\subseteq B\in\mathcal T$ $\Rightarrow$ $B\in\mathcal F$. (The definition of an open filter is similar to the definition of a filter, the only difference is that we are working only with open subsets of $X$.)
A maximal open filter is called an open ultrafilter. We can show, using Zorn lemma, that every open filter is contained in an open ultrafilter. In fact, we can show using Zorn lemma that every system of open sets, which has finite intersection property, is contained in an open ultrafilter.
If I am not missing something, then we can show that for an open ultrafilter $\mathcal F$ we have $\bigcap \mathcal F\ne\emptyset$ if and only if $\mathcal F$ converges.
Let us denote by $\mathcal N_x$ the system of all neighborhoods of $X$.
Suppose that $x\in\bigcap \mathcal F$. Clearly the system $\mathcal N_x\cup\mathcal F$ than has finite intersection property. (Any set from this system contains $x$.) Maximality of $\mathcal F$ then implies $\mathcal N_x\subseteq\mathcal F$, which means that $\mathcal F$ converges to $x$.
On the other hand, if $\mathcal F$ converges, then there exists a point $x\in X$ such that $\mathcal N_x\subseteq\mathcal F$.
This clearly implies $x\in\bigcap\mathcal F$. EDIT: As the counterexample posted in the answer shows, this is a place where my argument contained a mistake. In this way I can only show: An open ultrafilter converges if and only if it has a cluster point. (Or: An open ultrafilter $\mathcal F$ converges to $x$ if and only if $x$ is a cluster point of $\mathcal F$.)
The reason I have some doubts about this that I have seen mentioned in some papers that an open ultrafilter converges if and only if it has a cluster point.1 I thought that it would be quite logical to mention also the observation that it converges if and only if it has a non-empty intersection.
EDIT 2: My original claim is not true for Hausdorff spaces, either. It suffices to take any H-closed space (for example, any compact Hausdorff space) and choose any system of open subsets of $X$, which has finite intersection property, but has empty intersection. Using Zorn Lemma we can show that there is an open ultrafilter $\mathcal F$ containing this system. For this open ultrafilter we have $\bigcap \mathcal F = \emptyset$. But, since we are working in a H-closed space, the open ultrafilter $\mathcal F$ converges.
(A Hausdorff space $X$ is called H-closed if or absolutely closed if it is closed in any Hausdorff space, which contains $X$ as a subspace. It is known that a Hausdorff space is H-closed if and only if every open ultrafilter on $X$ converges.2)
More concrete example was given in Daniel Fischer's comment below.