I'll sketch the construction. It's inspired by the examples that Herrlich refers to, which partly go back to examples like the Tychonoff corkscrew (which is in Counterexamples in Topology) as an example of a regular space with only constant real-valued continuous functions.
Here we want constant $Y$-valued maps instead of real-valued ones (where $Y$ is our given $T_1$ space; I'll call a continuous $f:X \to Y$ a $Y$-valued map for short), and very similar "counter example machines" can be made (this is a construction where you input a space with a certain property $P$ and get a new space based on that space with some (hopefully better) property $P'$).
I'll start by a trivial remark that Herrlich doesn't even bother with initially (the final remark ("Bemerkung") does touch upon it): if $Y$ is finite, we're done right away: a finite $T_1$ space is discrete and we can take any large enough regular connected space $X$ (satisfying any of your cardinality demands, say a large power of $\mathbb{R}$ if need be, or any large enough infinite cofinite space) and note that any $Y$ valued map from a connected space to a discrete space is constant. So we can assume right from the beginning that $Y$ is infinite.
In fact, if we follow the construction, all we use of $Y$ is it's $T_1$-ness and its size (there isn't much more to know in general), say $|Y|=\aleph_\alpha$ for some ordinal $\alpha$. (Yes, some minor set theory knowledge is required and yes, I use AC but I don't mind that):
Step 1
Let $R_1$ be a set of size $\aleph_{\alpha+1}$, fix some $r_1 \in R_1$ and define the following variant of the one-point compactification of a discrete space: $R_1$ gets the topology where $R-\{r_1\}$ (I'll use $-$ for set difference, as Herrlich does) is discrete and a neighbourhood of $r_1$ is of the form $\{r_1\} \cup A$ where $|R_1-A| <|R_1|$ (so "small complements").
As $\aleph_{\alpha+1}$ has the property that the union of $\le \aleph_{\alpha}$ many subsets of size $< \aleph_{\alpha+1}$ (i.e. $\le \aleph_{\alpha}$) has size $<\aleph_{\alpha+1}$ as well, any intersection of $\aleph_{\alpha}$ many neighbourhoods of $r_1$ is still a neighbourhood of $r_1$.
Then we know that
(1a) any $Y$-valued map on $R_1$ is constant on a neighbourhood of $r_1$.
This follows from the set theory considerations: suppose $f: R_1 \to Y$ is continuous and let $y_0=f(r_1)$ and note that $$f^{-1}[\{y_0\}]= \bigcap \{ f^{-1}[Y-\{y\}]: y \in Y, y \neq y_0\}$$ which is (by continuity and $T_1$-ness) the intersection of $\aleph_{\alpha}$ many open neighbourhoods of $r_1$ and so the required open neighbourhood of $r_1$.
But we need another space $R_2$ with special point $r_2$: we do the exact same construction with $R_2, r_2 \in R_2$ where $|R_2|= \aleph_{\alpha+2}$. We again use sets of complement of size $<|R_2|$ for $R_2$ neighbourhoods and the rest is discrete.
This has the same property
(1b) any $Y$-valued map on $R_2$ is constant on a neighbourhood of $r_2$.
Bbut now $|R_1|$ is small relative to $|R_2|$ (which is used in the next step) so that
(1c) The intersection of $|R_1|$ many open neighbourhoods of $r_2$ is still a neighbourhood of $r_2$.
Also, we can easily check that both $R_1$ and $R_2$ are $T_1$ (finite sets are always closed because cofinite sets have a small complement by definition) and $T_3$ because there is only one non-isolated point in the space. (easy exercise, try it). And $|R_1|$ and $R_2|$ are both stricly larger than $|Y|\times |\mathbb N|$ ($=|Y|$, the $|\Bbb N|$ is only relevant if $Y$ were finite), and all other constructions will only increase cardinalities so cardinality will no longer be a concern.
Step 2
Now we make a "deleted plank" space, a very common idea (i.e. see the famous Tychonoff plank and its variations): $T=R_1 \times R_2 - \{(r_1,r_2)\}$ which is regular (in the subspace topology from the product topology) as $T_3$-ness is preserved by those operations. Now
(2) If $f$ is a $Y$-valued map on $T$ there are neighbourhoods $U_1$ of $r_1$ and $U_2$ of $r_2$ such that $f$ is constant on the open set $U_1 \times U_2 - \{(r_1,r_2)\}$.
Let such $f$ be given and note that for each $x \in R_1$ we can find an open neighbourhood $U_x$ of $r_2$ such that $f[\{x\} \times U_x]=\{f(x,r_2)\}$ by applying 1b to the map $f_x:t \to f(x,t): R_2 \to Y$. Then by 1c we know that $U_2 = \bigcap_{x \in R_1, x \neq r_1} U_x$ is an open neighbourhood of $r_2$. Then let $y\in V_2, y\neq r_2$, and we apply 1a to find an open neighbourhood $U_1$ of $r_1$ such that $f[U_1 \times \{y\}]= f(r_1,y)$. It now follows that $f$ is constant on $U_1 \times U_2 - \{(r_1,r_2)\}$, as required.
Step 3
This is similar to the ideas in Hewitt's Condensed Corkscrew, or the Tychonoff Corkscrew as described in "Counterexamples".
We take countably many copies of $T$, call them $T_n$, and to their topological sum space we add two new points $a$ and $b$ that have as basic neighbourhoods $\{a\} \cup \bigcup_{k \le n} T_k$ (where $n$ varies over the natural numbers) and $\{b\} \cup \bigcup_{k > n} T_k$ (same $n$'s), respectively. This sum space plus extra points is still $T_3$ of course and then Herrlich defines a quotient space $Q$ of it, by identifying:
for all even $n$ all points $(x,r_2)$, $x \in R_1, x \neq r_1$ in $T_{n}$ with the corresponding $(x,r_2)$ in $T_{n+1}$, and
for all odd $n$ all points $(r_1,x)$, $x \in R_2, x \neq r_2$ in $T_n$ with the corresponding $(r_1,x)$ in $T_{n+1}$.
The resulting quotient space $Q$ has the property
(3) If $f$ is a $Y$-valued map on $Q$, then $f(a)=f(b)$
For a proof (if you cannot figure it out) check the references of Herrlich's paper (Hewitt or Novak maybe) or check examples 90-94 in Counterexamples, e.g.
The paper just makes the claim without proof (though it should intuitively make sense, as a $Y$-valued map on each $T_n$ is constant on a large "corner rectangle", and these are all glued together by the identification, thereby pulling $a$ and $b$ along to those constant values...).
Note that we're getting closer to the required space, we can start a machine now
that uses the fact (3) for two special points in the starting and using two more constructions we get a space that has the property more globally.
Step 4
Now for any space $Z$ we can make a space $Q(Z)$, containing $Z$ as a subspace, with the following property:
(4) If $f$ is a $Y$-valued map on $Q(Z)$ then $f$ is constant on $Z$.
Another nice improvement on the situation with two points $a$ and $b$ that we have for $Q$.
We start by forming the Cartesian product $Z \times Q$ and giving points of it basic neighbourhoods as follows: Points of the form $(z,x)$ where $x \neq a$ have open neighbourhoods of the form $\{z\} \times U$ where $U$ is an open neighbourhood of $x$ in $Q$, while points of the form $(x,a)$ have neighbourhoods of the form $U \times \{a\}$, where $U$ is an open neighbourhood of $x$. So this is a kind of assymetric product space (where we use vertical or horizontal stalks instead of squares). Now $Q(Z)$ is defined as the quotient of the above under the identification where the subset $Z \times \{b\}$ is identified to a single point in $Q(Z)$. $Z$ can be seen as a subspace of $Q(Z)$ via the map that sends $z$ to the class of $(z,a)$. You can also check that if $Z$ is $T_3$ then so is $Q(Z)$, so claims Herrlich. (probably just some tedious case checking) and also the assertion (4) from above is not proved in detail. Again the references will contain similar arguments, probably.
Step 5
The final space $X$ with only constant $Y$-valued maps can now be constructed:
First set $X_0=\{p\}$ for some point $p$ and define by recursion $X_{n+1}=Q(X_n)$, applying the construction from step 4 (which depended on the space $Q$, which was built from copies of $T$, etc.) and on $X:=\bigcup_{n \in \Bbb N} X_n$ we put the "inductive" topology (recall that we have $X_n \subseteq X_{n+1}$ for all $n$, so we have an increasing union of spaces): $O$ is open in $X$ iff $O \cap X_n$ is open in $X_n$ for each $n$. (This is a standard inductive limit construction in many parts of topology, in fact.)
Now if $f: X \to Y$ is continuous and $x, y$ in $X$, then $x \in X_n$, $y\in X_m$ for some $n,m \in \Bbb N$ and then $x,y \in X_k$ with $k = \max(n,m)$ and as $f|_{X_{k+1}}$ is continuous, and $Y$-valued and so $f$ is constant on $X_k$ (by (4) and $X_{k+1}=Q(X_k)$ and $f(x)=f(y)$. So $f$ is constant on $X$.
Finally see also this thread, which has a link to an English translation of Herrlich's paper, by Martin Sleziak (also active on this site) and the answers contain more links and references. Brian Scott's answer reconstructs a similar "machine" by van Douwen.
This thread (again some nice writeup by Brian Scott, relevant for step 3) also gives some proof ideas, perhaps.
