Largest Hausdorff quotient

Consider the equivalence relation $\sim$ on your space $X$ such that $x\sim y$ iff $x$ and $y$ have the same image under all surjective continuous maps $f:X\to Y$ with codomain $Y$ a Hausdorff space. Put on the set $X/\sim$ the least topology which makes all those maps continuous, and you have the space you want. I doubt there is any actual reference for this.

Mariano already answered the question, but let me make two additional remarks:

1) Actually the proof of GAFT is constructive and can be (at least sometimes) used to get explicitly a left adjoint functor. In this case, you can see directly the "big coequalizer" whose set-theoretic existence issue is dealt with the solution set condition: just consider all maps from your space $X$ into Hausdorff spaces which are surjective; this bounds the cardinality of the Hausdorff spaces, and thus up to isomorphism there is only a set of them.

2) Another construction of the left adjoint to $\mathsf{Haus} \to \mathsf{Top}$ works as follows: Let $X$ be a topological space, and consider the equivalence relation $\sim$ generated by: If $x,y$ cannot be separated by disjoint open sets, then $x \sim y$. Then $H(X):=X / \sim$ has the property that every map from $X$ into a Hausdorff space uniquely factors through $X \to H(X)$. If $H(X)$ was Hausdorff, we would be done. But this is not always the case. Instead, we have to repeat this construction: $X \to H(X) \to H(H(X)) \to H(H(H(X))) \to \dotsc$, then take the colimit $H^{\omega}(X)$, and make again $H^{\omega}(X) \to H(H^{\omega}(X)) \to \dotsc ...$. You can continue this for every ordinal number. Since $X$ is a set and all these maps are quotient maps, at some stage we get an isomorphism, which is the desired Hausdorff quotient.

It is interesting when we arrive at this stage, see my question about the nonhausdorff dimension.

To augment very slightly Mariano's nice answer, Hausdorff quotients (as opposed to surjections) suffice.

To obtain the finest Hausdorff quotient of an arbitrary space $X$, take the quotient of $X$ by the intersection of all partitions of $X$ determined by quotient maps from $X$ with $T_{2}$ image.