As I point out in a comment, these questions came to mind while I was doing a related exercise in Munkres' "Topology". I am pleased to see that it is a problem in the Magazine!
I will give here a partial answer that does not affect the problem in the magazine.
Statement: For any given $n\ge 3$ there is a finite topological space containing a 14-set of cardinality $n$.
Proof: For $n\ge 4$, let $m=n+3$, the point set $X=\{1,2,3,\ldots, m\}$, and $\mathcal{B}=\{\emptyset, X, \{1\}, \{6\}, \{1,2\}, \{3,4\}, \{5,6\}\}$ as the basis for the topology on $X$. Then the subset $A=\{1,3,5\}$ is a 14-set of cardinality 3. Therefore, $A'$ (the complement of $A$ in $X$) is a 14-set of cardinality $m-3=n$, as desired.
For $n=3$, let $m=7$ in the previous setting and $A$ is the desired 14-set.
Moreover, we can obtain 14-sets of any desired infinite cardinality as follows: Let $Y$ be a set of cardinality $k\ge \aleph_{0}$, and let $X=\{1,2,3,4,5,6,7\}\cup Y$ with basis for the topology $\mathcal{B}$ as before. Then the set $A$ is a 3-element 14-set in $X$, and $A'$ is a 14-set of cardinality $k$. For example, if $Y=\mathbb{N}$ then $A'$ is a countably infinite 14-set.
Answer added Nov. 18:
The smallest possible cardinality of a Kuratowski 14-set $E$ is $3$. Let $k$, and $i$ denote closure and interior, respectively. Then $E$ is a Kuratowski 14-set iff $E$, $E^{i}$, $E^{iki}$, $E^{ki}$, $E^{ik}$, $E^{kik}$, and $E^{k}$ are all non-empty, distinct sets strictly contained in $X$. If $E$ is empty the answer is trivial. If $|E|=1$ then $E^{i}$ equals either $E$ or is empty. If $E=\{a,b\}$ then we may assume that $E^{i}=\{a\}$. But then $a\notin B=E^{ki}-E^{ik}$, which is an open set contained in $E^{k}$ and so it must intersect $E$ non-trivially. Hence, $b\in B$ from which we obtain $E^{kik}=E^{k}$.
Remark: Since a Kuratowski 14-set $E$ of cardinality 2 or less does not exist, we may ask what is the maximum number of different sets obtainable from $E$ by taking closures and complements.
If $|E|=1$ we have two possibilities:
First, if $E^{i}=E$ then $E^{ik}=E^{kik}=E^{k}$ and $E^{iki}=E^{ki}$. Therefore, at most 6 distinct sets can be obtained from $E$ by taking closures and complements. This bound is possible if $|X|\ge 3$. For example, $X=\{1,2,3\}$ with basis $\{\{3\},X\}$ and $E=\{1\}$.
Second, if $E^{i}=\emptyset$ then $E^{ik}=E^{iki}=\emptyset$ and $E^{kik}=E^{k}$. Therefore, at most $8$ distinct sets can be obtained from $E$ by taking closures and complements. This bound is possible if $|X|\ge 4$. For example, $X=\{1,2,3,4\}$ with basis $\{\{4\},\{1,2\},X\}$ and $E=\{1\}$.
If $E=\{a,b\}$ we may assume that $E^{i}=\{a\}$. But then $a\notin B=E^{ki}-E^{ik}$, which is an open set contained in $E^{k}$ and so it must intersect $E$ non-trivially. Hence, $b\in B$ from which we obtain $E^{kik}=E^{k}$. Therefore, at most 12 distinct sets can be obtained from $E$ by taking closures and complements. This bound is possible if $|X|\ge 6$. For example $X=\{1,2,3,4,5,6\}$ with basis $\{\{1\},\{6\},\{1,2\},\{4,5\},X\}$ and $E=\{1,4\}$.