Is there a simple formula in the language of set theory that defines cardinality for finite sets without using Choice?
From here, it appears that the full cardinality function in ZFC can be defined with
$$|A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}.$$
Is this function well-defined for finite sets without choice, say in ZF\Inf? Is there a simpler way to define the cardinality function for finite sets?
Finite, as defined normally, means "in bijection with a finite ordinal". You can argue that "finite ordinal" require us to define finite first, but we can also define an ordinal to be a "finite ordinal" if it is $0$ or a successor ordinal and any smaller ordinal is $0$ or a successor.
This means that if $A$ is finite, it is in bijection with an ordinal. So the class we are trying to choose from is never empty, so the choice is well-defined.
The axiom of infinity, while necessary for proving that an infinite ordinal exists, is not needed here either.
Indeed, the only place where choice is ever used is when you want to argue that for any given set $A$, the class of ordinals is non-empty, so we can pick its minimum (which does not require choice, since the minimum is well-defined as long as the class is non-empty).
You may ask, then, how do we define the cardinal of a set which cannot be well-ordered, in which case I will first point out that finite sets are by definition well-orderable, so it is irrelevant, but to give an idea, we have the von Neumann hierarchy and we can use it to "cut out" a set from $\{B\mid\exists f\colon A\to B\text{ a bijection}\}$ in a uniform way and use that to represent the cardinality of $A$. But, again, this is not relevant for well-orderable (and in particular finite) sets.
