While johndoe gave the right answer to the question I believe you meant to ask, the question that you did ask is slightly different and also has a negative answer. Specifically, johndoe's answer addresses the version of your question in which the factorization $\bar{f}$ is required to be unique. However, even if you do not require the factorization to be unique, then a co-Hausdorffification still does exist. (One might call such a thing a weak co-Hausdorffification, by analogy with the term weak limit.)
Specifically, let $X=\{0,1\}$ be the Sierpinski 2-point space (with $1$ closed), and suppose $f:T\to X$ is a map from a Hausdorff space to $X$. Then I claim that there exists a Hausdorff space $Z$ and a map $g:Z\to X$ that does not factor through $f$. To show this, let $\kappa$ be any ordinal of cofinality greater than $|T|$, let $K=\kappa+1$, and let $g:K\to X$ send $\kappa$ to $1$ and everything else to $0$. Suppose $h:K\to T$ is such that $fh=g$. Since $\operatorname{cf}(\kappa)>|T|$, there is an unbounded set $S\subseteq\kappa$ on which $h$ is constant. By continuity of $h$, we must then have $h(\alpha)=h(\kappa)$ for all $\alpha\in S$. But then $g(\alpha)=f(h(\alpha))=f(h(\kappa))=g(\kappa)$ for all $\alpha\in S$, a contradiction.
(In the language of the adjoint functor theorem, this is saying that a co-Hausdorffification fails to exist not only because Hausdorff spaces are not closed under coequalizers, but also because the solution set condition fails.)