if $X$ is a topological space, a first step in making $X$ hausdorff is taking the quotient $H(X)=X/\sim$, where $\sim$ is the equivalence relation generated by the topological indistinguishable points (i.e.: if $x,y$ cannot be seperated by disjoint open sets, then $x \sim y$). observe that $X$ is hausdorff, when $X \to H(X)$ is an isomorphism, and that for every hausdorff space $K$ the map $Hom(H(X),K) \to Hom(X,K)$ induced by the projection $X \to H(X)$ is bijective.
by a fairly general categorical argument, we can construct from this the free functor from topological spaces to hausdorff spaces (i.e. it's left adjoint to the forgetful functor): for ordinal numbers $\alpha$, define the functor $H^\alpha$ (together with natural transformations $H^{\alpha} \to H^{\beta}, \alpha < \beta$) by $H^0 = id, H^{\alpha+1} = H \circ H^\alpha$ and $H^\alpha = colim_{\delta < \alpha} H^\delta$. for every topological space $X$ there is an ordinal number $\alpha$ such that $H^\alpha(X) = H^{\alpha+1}(X)$, then $H^\alpha(X)$ is the free hausdorff space associated to $X$. define $h(X)$, the "nonhausdorff dimension" to be the smallest such ordinal number $\alpha$. every ordinal number arises as a nonhausdorff dimension(!).
I've came up with this with a friend and we don't know of any literature about it. perhaps someone of you has already seen it elsewhere? there are some further questions: every $H^\alpha(X)$ is a quotient of $X$, but how can we describe the equivalence relation explicitely? what is the intuition for a space $X$ to have nonhausdorff dimension $\alpha$? are there known classes of topological spaces whose nonhausdorff dimension can be bounded? and of course: is there some use for the nonhausdorff dimension? ;-)