The Sierpiński space, $S=\{0,1\}$ with the topology $\{ \{\}, \, \{1\}, \, \{0,1\}\}$, is a $T_{0}$ space. And in some sense, it is the "prototypical" $T_{0}$ space, for, if $(X,\tau)$ is a topological space, $X$ is $T_{0}$ iff $X$ can be seen as a subspace of $\prod _{A \in \tau} S$. To be accurate, $X$ is $T_{0}$ if, and only if, $$ X \hookrightarrow \prod _{A \in \tau} S \; | \; x \mapsto (\mathbb{I} _{A}(x))_{A \in \tau} $$ (I mean, the arrow is injective, and the topology on $X$ is the same as the initial topology with respect to this arrow.)
Are there similar characterizations of other kinds of spaces? Are there similar exemplary $T_{q}$ spaces for other $q$?
As mentioned in the comments, $[0,1]$ is an "exemplary" completely regular space; this is essentially the definition of "completely regular". On the other hand, there is no exemplary $T_1$ or $T_2$ space.
For $T_1$ spaces, let $S$ be any infinite set and endow $S$ with the cofinite topology. Note that if $f:S\to X$ is a continuous map from $S$ to a $T_1$ space with $|X|<|S|$, then $f$ must be constant: since $|X|<|S|$, there must be some $x\in X$ such that $f^{-1}(\{x\})$ is infinite, but then $f^{-1}(\{x\})$ must be all of $S$ since it must be closed. It follows that if $X$ is an exemplary $T_1$ space, it must have at least $|S|$ points. Since $S$ is arbitrary, this shows that no exemplary $T_1$ space exists.
For $T_2$ spaces, let $S$ be any infinite set with the discrete topology, let $S\cup\{\infty\}$ be its $1$-point compactification, and consider the space $X=S\times (S\cup\{\infty\})\cup\{z\}$, topologized as follows. The subset $S\times (S\cup\{\infty\})$ is an open subset of $X$ and has the usual product topology. A set $U\subseteq X$ containing $z$ is open iff for all but finitely many $s\in S$, $U$ contains all but finitely many points of $\{s\}\times S$. It is easy to see this space $X$ is $T_2$. Now suppose $Y$ is any $T_2$ (or even just $T_1$) space and $f:X\to Y$ is continuous. Then for each $s\in S$, there is a $y_s\in Y$ such that $f(s,t)=y_s$ for all but $|Y|$ many $t\in S$ (otherwise, there would be two different values which $f(s,t)$ takes for infinitely many $t$, and then $f(s,\infty)$ would have to be equal to both of them). Similarly, there is then a single $y\in Y$ such that $y_s=y$ for all but $|Y|$ many $s\in S$ (otherwise two values of $y_s$ would occur infinitely often, and then $f(z)$ would have to be equal to both of them). We then have $f(z)=y$ and $f(s,\infty)=y$ for all but $|Y|$ many values of $s$. It follows that if $U\subseteq Y$ is any open set such that $z\in f^{-1}(U)$, $(s,\infty)\in f^{-1}(U)$ for all but $|Y|$ many values of $s$. If $|Y|<|S|$, this means that if $i:X\to \prod Y$ is any continuous injection from $X$ to a product of copies of $Y$, the set $i(X\setminus (S\times\{\infty\}))$ cannot be open in the subspace topology on $i(X)$ (since any neighborhood of $i(z)$ has to contain all but $|Y|$ many points of $S\times \{\infty\}$). Since $X\setminus (S\times\{\infty\})$ is open in $X$, this means $Y$ cannot be exemplary for $X$. Thus any exemplary $T_2$ space must have at least $|S|$ points; since $S$ was arbitrary, this means there is no exemplary $T_2$ space.