Terminology for the cardinality of the transitive closure of a set. Let us say that a set $X$ is transitive if for all $x \in X$, if $x$ is a set then $x \subseteq X$. 
(This is all from a naive set theory perspective; I'm assuming, for instance, that $\{1, 2, 3\}$ is a transitive set because none of its elements are sets.)
Now, let us define the transitive closure $T(X)$ of a set $X$ to be the intersection of all transitive sets of which $X$ is a subset. Roughly speaking, the transitive closure of $X$ is $X \cup$ the elements of the elements of $X \cup $ the elements of the elements of the elements of $X$, and so on.
If $T(X)$ is a finite set, then obviously $X$ is a finite set. Moreover, there's a sense in which $X$ is "definitely" a finite set, in a way that e.g. $\{\mathbb{N}\}$ is not, since even though $|\{\mathbb{N}\}| = 1$, we have $|T(\{\mathbb{N}\})| = |\{\mathbb{N}\} \cup \mathbb{N}| = \aleph_0$.
With that in mind, is there a standard name (analogous to "the cardinality of $X$") for the cardinality of $T(X)$?
Is there a standard term (analogous to "X is a finite set") for sets $X$ such that $T(X)$ is a finite set? 
 A: I didn't hear any name of the cardinality of $T(x)$. It is just usually called the cardinality of the transitive closure of $x$. However, if $T(x)$ is countable, $x$ is often called a hereditary countable set. 
Now I am going to answer your second question:
Claim. $T(x)$ is finite if and only if $x\in V_\omega$, where $V_\omega$ is the $\omega$th von Neumann hierarchy.
Proof. Note that $V_n$ is transitive for all $n$. If $x\in V_\omega$, then there is some $n<\omega$ such that $x\in V_n$. Since $V_n$ is transitive and $x\in V_n$, $x\subseteq V_n$ and $T(x)\subseteq V_n$, by your definition of the transitive closure. Therefore $T(x)$ is finite.
The proof of the converse uses the induction on the size of $T(x)$: assume that if $|T(y)|<n$ then $y\in V_\omega$. Now assume that $|T(x)|=n$. Note that
$T(x) = x\cup T\left(\bigcup x\right)$ and $x$ is non-empty so $|T(\bigcup x)|< n$ and $T(\bigcup x)\in V_\omega$. From this we have $x\subseteq V_\omega$.
Since $x$ is finite, every element of $x$ lies on for some $V_m$, so $x\in V_{m+1}\subseteq V_\omega$. You can check that $V_\omega$ is closed under finite union therefore $T(x)\in V_\omega$.
This is the reason why the set $V_\omega$ is called the set of hereditary finite sets. 

Added on 23 April 2020: Someone asked to me that $|T(\bigcup x)|< n$ in the above proof holds. Here is the proof of it:
We know that $T(x) = x\cup T\left(\bigcup x\right)$. Hence it suffice to prove that $x\nsubseteq T(\bigcup x)$. Since $x$ is finite, we can choose $y\in x$ whose rank is greatest among that of elements of $x$. If $w\in z\in x$, then $\operatorname{rank}w <\operatorname{rank}z\le\operatorname{rank}y$. Hence $y\notin \bigcup x$. By the same argument, we have $y\notin T(\bigcup x)$.
