A free group in what category/variety?
Let $\mathcal{C}$ be a class of groups. We say that a group $F\in\mathcal{C}$ is a free $\mathcal{C}$-group if there is a subset $X\subseteq F$ such that for every group $G\in\mathcal{C}$ and every function $f\colon X\to G$, there is a unique group homomorphism $\mathfrak{f}\colon F\to G$ such that $\mathfrak{f}(x)=f(x)$ for all $x\in X$.
When we talk about "free groups", with no qualifications, we are generally talking about the "absolutely free groups", which are the ones where $\mathcal{C}$ is the class of all groups. Another common class is when $\mathcal{C}$ is the class of all abelian groups, in which case the "free $\mathcal{C}$-group" are the "free abelian groups."
It is easy to see that $\mathbb{Z}_4$ cannot be a free group or a free abelian group. To see this: if $X\subseteq\mathbb{Z}_4$ is not empty, with $x\in X$, then no map that sends $x$ to $1\in\mathbb{Z}$ can be extended to a group homomorphism from $\mathbb{Z}_4$ to $\mathbb{Z}$ (all such homomorphism map everything to $0$). And if $X=\varnothing$, then there would have to be a unique group map from $\mathbb{Z}_4$ into any abelian group, and this is not the case: there are, for example, four different homomorphisms $\mathbb{Z}_4\to\mathbb{Z}_4$.
For the same reason, since it is not a free abelian group, it cannot be an absolutely free group either.
Those are the two most reasonable interpretations of the question.
That said, there is a "nice" class of groups for which $\mathbb{Z}_4$ is a $\mathcal{C}$-free group: the class of abelian groups of exponent $4$. Take $X=\{1\}$. If $G$ is any abelian group of exponent $4$, and $f\colon\{1\}\to G$ is any function, then this determines a unique homomorphism $\mathbb{Z}_4\to G$, which works because $4f(1)=0$ in $G$.