Fix a space $X$. Below we will introduce a sequence of spaces associated to it. I will understand the terms regular and completely regular to not include the $T_1$ axiom, and each subcategory I introduce will be full and replete.
Firstly denote by $hX$ the maximal Hausdorff quotient of $X$. This is a quotient space of $X$ with the property that any map $f:X\rightarrow Y$ from $X$ into a Hausdorff space factors uniquely through a map $f':hX\rightarrow Y$. I'll link those interested in a definition to this post. In any case, the assignment $X\mapsto hX$ is functorial and defines a functor $Top\rightarrow Top_2$ from the category of all topological spaces into the category of Hausdorff spaces.
Secondly we will write $crX$ for the completely regular modification of $X$. To define it notice that the cozero sets in $X$ form a base for a topology on the underlying set which is weaker than $X$'s existing topology. Then $crX$ is the underlying set of $X$ equipped with this new topology. The identity induces a continuous map $X\rightarrow crX$, and any map $f:X\rightarrow Y$ from $X$ into a completely regular space $Y$ factors uniquely through a map $f'':crX\rightarrow Y$. Again this gives a functor $Top\rightarrow Top_{CR}$, $X\mapsto crX$, now into the category $Top_{CR}$ of completely regular spaces.
Notice that $crX$ is Hausdorff if and only if $X$ is completely Hausdorff. In general $crX$ will not be Hausdorff, even if $X$ itself is. Similarly $hX$, although Hausdorff, need not be completely regular. Thus we are led to consider the spaces $h(crX)$ and $cr(hX)$. Now the first of these spaces need not be completely regular, and the second need not be Hausdorff, so we introduce the spaces $cr(h(crX))$ and $h(cr(hX))$. Continuing on we obtain a potentially infinite sequence of spaces, obtained by alternative application of the functors $h$ and $cr$. The sequence will terminate exactly when we apply $h$ to a Hausdorff space, or $cr$ to a completely regular space.
Do these sequences necessarily converge to completely regular Hausdorff spaces?
I'm guessing that they do not. For instance, if we replace $h$ with the k-ification functor, which replaces $X$ with its compactly generated reflection $kX$, then it is know that there is a completely regular space $Y$ and a k-space $Z$ for which $kY=Z$ and $crZ=Y$. On the other hand I have no counterexample to my question above.
Note: $X$ always has its completely regular Hausdorff, aka Tychonoff, modification $tX$, which is the image of the canonical map $X\rightarrow I^{C(X,I)}$. Although $tX$ has a universal property with respect to Tychonoff spaces, this is not the space I am looking for. I am interested in the sequence I construct above.
In fact, $h(crX)$ is always completely regular and is the same as $tX$. To see this, note that $crX$ is just the coarsest topology on $X$ that makes the canonical map $X\to I^{C(X,I)}$ continuous. Since the image of this map is $tX$, the induced map $crX\to tX$ is a surjection with the property that the domain has the coarsest topology that makes it continuous. In other words, this means that points of $X$ which map to the same point of $tX$ are topologically indistinguishable in $crX$, and so $crX\to tX$ is just the $T_0$-ification of $crX$. Since $tX$ is in fact not just $T_0$ but Hausdorff, $crX\to tX$ is also the Hausdorffification of $crX$.
(To put it another way, what's going on here is that completely regular spaces are very close to being the same as Tychonoff spaces--they are just the spaces whose $T_0$-ification is Tychonoff.)