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 not exist.  (One might call such a thing a weak co-Hausdorffification, by analogy with the term [weak limit](http://ncatlab.org/nlab/show/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.)