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.)
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. Nevertheless, it turns out that the (large) colimit of all Hausdorff spaces mapping to $X$ does exist in the category of all spaces: it is just $X$ itself. This is not hard to prove; it essentially amounts to showing that any topology is determined by which maps from Hausdorff spaces are continuous.)