It seems that there is no such space. Indeed, I claim that every point in a Boolean space without isolated points is the limit of a long sequence in that space.

We may assume that the space $X$ is the Stone space of an atomless Boolean algebra $B$, so that the points in $X$ consist of all the ultrafilters in $B$, and the basic open sets have the form $\{U\in X\mid b\in U\}$, where $b\in B$.

I claim that every point $U$ in $X$ is a limit of a long sequence. To see this, let $\langle b_\alpha\mid\alpha\lt\delta\rangle$ be a well-ordered enumeration of a minimal sequence of generators of $U$ as an ultrafilter in $B$. That is, $U$ is the filter generated by the $b_\alpha$'s, but it is not generated by any initial segment of this sequence. For each $\gamma\lt\delta$, let $F_\gamma$ be the filter generated by $\{b_\alpha\mid\alpha\lt\gamma\}$. Since this initial segment of the generators does not generate all of $U$, we may extend this filter $F_\gamma$ to an ultrafilter $U_\gamma\in X$ distinct from $U$.

I claim that the long sequence $\langle U_\gamma\mid\gamma\lt\delta\rangle$ converges to $U$. Every basic open neighborhood of $U$ is determined by some $b\in U$, which must be in some $F_\beta$ for some $\beta\lt\delta$. It follows for $\beta\leq\gamma\lt\delta$ that we have $b\in F_\gamma\subset U_\gamma$, and so all such $U_\gamma$ are also in the basic open set determined by $b$. In other words, for any basic open set containing $U$, there is a tail of the sequence $U_\gamma$ that is inside it. So $U_\gamma\to U$, as desired.

The length of the sequence converging to $U$ is the same as the cardinality of the minimal generating set of $U$ as a filter.

It seems that there is no such space. Indeed, I claim that every non-isolated point in a Boolean space is the limit of a long sequence in that space.

We may assume that the space $X$ is the Stone space of a Boolean algebra $B$, so that the points in $X$ consist of all the ultrafilters in $B$, and the basic open sets have the form $\{U\in X\mid b\in U\}$, where $b\in B$.

I claim that every non-isolated point $U$ in $X$ is a limit of a long sequence. To see this, let $\langle b_\alpha\mid\alpha\lt\delta\rangle$ be a well-ordered enumeration of a minimal sequence of generators of $U$ as an ultrafilter in $B$. That is, $U$ is the filter generated by the $b_\alpha$'s, but it is not generated by any initial segment of this sequence. Since $U$ is not isolated, it follows that $\delta$ must be infinite, and by rearranging if necessary, we may assume that $\delta$ is a limit ordinal. For each $\gamma\lt\delta$, let $F_\gamma$ be the filter generated by $\{b_\alpha\mid\alpha\lt\gamma\}$. Since this initial segment of the generators does not generate all of $U$, we may extend this filter $F_\gamma$ to an ultrafilter $U_\gamma\in X$ distinct from $U$.

I claim that the long sequence $\langle U_\gamma\mid\gamma\lt\delta\rangle$ converges to $U$. Every basic open neighborhood of $U$ is determined by some $b\in U$, which must be in some $F_\beta$ for some $\beta\lt\delta$. It follows for $\beta\leq\gamma\lt\delta$ that we have $b\in F_\gamma\subset U_\gamma$, and so all such $U_\gamma$ are also in the basic open set determined by $b$. In other words, for any basic open set containing $U$, there is a tail of the sequence $U_\gamma$ that is inside it. So $U_\gamma\to U$, as desired.

The length of the sequence converging to $U$ is the same as the cardinality of the minimal generating set of $U$ as a filter.

It seems that there is no such space.

We may assume that $X$ is the Stone space of an atomless Boolean algebra $B$, so that the points in $X$ consist of all the ultrafilters in $B$, and the basic open sets have the form $\{U\in X\mid b\in U\}$, where $b\in B$.

I claim that every point $U$ in $X$ is a limit of a long sequence. To see this, let $\langle b_\alpha\mid\alpha\lt\delta\rangle$ enumerate a minimal sequence of generators of $U$ as an ultrafilter in $B$. That is, $U$ is the filter generated by the $b_\alpha$'s, but it is not generated by any initial segment of this sequence. For each $\gamma\lt\delta$, let $F_\gamma$ be the filter generated by $\{b_\alpha\mid\alpha\lt\gamma\}$. Since this initial segment of the generators does not generate all of $U$, we may extend this filter $F_\gamma$ to an ultrafilter $U_\gamma\in X$ distinct from $U$.

I claim that the long sequence $\langle U_\gamma\mid\gamma\lt\delta\rangle$ converges to $U$. Every basic open neighborhood of $U$ is determined by some $b\in U$, which must be in some $F_\beta$ for some $\beta\lt\delta$. It follows for $\beta\leq\gamma\lt\delta$ that we have $b\in F_\gamma\subset U_\gamma$, and so all such $U_\gamma$ are also in the basic open set determined by $b$. In other words, for any basic open set containing $U$, there is a tail of the sequence $U_\gamma$ that is inside it. So $U_\gamma\to U$, as desired.

The length of the sequence converging to $U$ is the same as the cardinality of the minimal generating set of $U$ as a filter.

It seems that there is no such space. Indeed, I claim that every point in a Boolean space without isolated points is the limit of a long sequence in that space.

We may assume that the space $X$ is the Stone space of an atomless Boolean algebra $B$, so that the points in $X$ consist of all the ultrafilters in $B$, and the basic open sets have the form $\{U\in X\mid b\in U\}$, where $b\in B$.

I claim that every point $U$ in $X$ is a limit of a long sequence. To see this, let $\langle b_\alpha\mid\alpha\lt\delta\rangle$ be a well-ordered enumeration of a minimal sequence of generators of $U$ as an ultrafilter in $B$. That is, $U$ is the filter generated by the $b_\alpha$'s, but it is not generated by any initial segment of this sequence. For each $\gamma\lt\delta$, let $F_\gamma$ be the filter generated by $\{b_\alpha\mid\alpha\lt\gamma\}$. Since this initial segment of the generators does not generate all of $U$, we may extend this filter $F_\gamma$ to an ultrafilter $U_\gamma\in X$ distinct from $U$.

I claim that the long sequence $\langle U_\gamma\mid\gamma\lt\delta\rangle$ converges to $U$. Every basic open neighborhood of $U$ is determined by some $b\in U$, which must be in some $F_\beta$ for some $\beta\lt\delta$. It follows for $\beta\leq\gamma\lt\delta$ that we have $b\in F_\gamma\subset U_\gamma$, and so all such $U_\gamma$ are also in the basic open set determined by $b$. In other words, for any basic open set containing $U$, there is a tail of the sequence $U_\gamma$ that is inside it. So $U_\gamma\to U$, as desired.

The length of the sequence converging to $U$ is the same as the cardinality of the minimal generating set of $U$ as a filter.