Q:

How do set theorists view this issue?

Since the question has changed significantly over the course of last few days, it was suggested to modify it in that light. I have removed those parts which aren't directly related to the main question, as it currently stands now. The current question is mostly just what I wrote in the "Edit:" in the previous version, along with few simplifications and small corrections.
The statement $(**)_1$ was mentioned in the answer (by Noah Schweber). Suppose that a hierarchy such as that in $(**)_1$ exists. Now consider the plausibility argument below of $(**)_1$ being "possibly" true. 
Suppose that such a $\omega_1+1$ length hierarchy exists and that the $\omega_1$-point of such a hierarchy is the function $g:\mathbb{N} \rightarrow \mathbb{N}$. Now for any point $\alpha<\omega_1$ we denote the function at the point $\alpha$ as $f_\alpha:\mathbb{N} \rightarrow \mathbb{N}$. Write "$f_1$ eventually dominates $f_2$" as $f_1>f_2$. Now whenever we have $\alpha<\beta<\omega_1$, we also have $f_\alpha < f_\beta$. Furthermore we have $g>f_\alpha$ for all $\alpha<\omega_1$.
Now we want to define a  collection of sets $C_i$ (where $i \in \omega$) such that they partition all limit values below $\omega_1$. Meaning all the $C$'s are disjoint and their union is the set of all limit values below $\omega_1$. We define the set $C_n$ (for arbitrary $n$) in the following manner. A given limit element $\alpha<\omega_1$ belongs to $C_n$ if and only if:
$$g(n-1) \le f_\alpha(n-1)$$ 
$$g(x) > f_\alpha(x) \,\,\,\, for \,\, all \,\, x\ge n$$ 
For the definition of $C_0$ we drop the line $g(n-1) \le f_\alpha(n-1)$ and just consider the second line.
Now we want to consider two separate scenarios.
(1) 
For all values $i \in \omega$ the least upper bound of values in $C_i$ is strictly less than $\omega_1$. Symbolically we write, for all $i \in \omega$ we have $sup(C_i)<\omega_1$.
In this case loop through limit values from $\omega$ to $\omega_1$ and mark whether they belong to $C_0$ or not. Mark the least upper bound of elements in $C_0$ as $a_0$ (this is first term of sequence). Do this again for $C_1$ and mark the sup of values in $C_1$ as $a_1$. if it turns out that $a_1 \le a_0$ then ignore $a_1$. Otherwise set the second term of sequence to $a_1$. Repeat this all the way $\omega$ number of times. 
This gives a sequence for $\omega_1$. This leaves the possibility for scenario (2) which is:  
(2) There exist some values $i \in \omega$ such that $sup(C_i)=\omega_1$
Specifically let's just assume that $sup(C_0)=\omega_1$. The definitions below can be easily generalised for any value $k \ne 0$.
Define a function $h^0_\alpha:\mathbb{N} \rightarrow \mathbb{N}$. Denote the smallest element of $C_0$ as $p_0$. We define:
$$h^0_\alpha(x)=0 \quad for \,\,\, \alpha \le p_0$$
For $\alpha>p_0$ we have: $h^0_\alpha(x)=max\{f_\beta(x)\,|\, \beta<\alpha$ and $\beta \in C_0 \}$. Now the function $h^0_\alpha$ can only change countable number of times and, furthermore, there must be a point $N_0<\omega_1$ after which it stops changing. So I guess whenever we have $sup(C_0)=\omega_1$ such a value $N_0<\omega_1$ would always exist? If it doesn't exist, then we have a contradiction. 
The consideration in the above paragraph applies to all values $k$ for which we have $sup(C_k)=\omega_1$. I also hope, if it is known, someone points to a reference for proof (or an argument) for the existence of $N_k$ whenever $sup(C_k)=\omega_1$. 

A:

What you've written is correct, but it doesn't end in a contradiction.
Here's an example of how $N_0$ can exist - and indeed be small - without really doing anything (I'm going to work with all ordinals, not just limit ordinals, since that's more natural - there's no reason in your current argument to restrict attention to limits):

Let $g(x)=x+1$.
For each $k\in\omega$, $f_k(x)=\min\{g(x)-1, k\}$.
Let $f_\omega=floor(\ln(x+1))$.

Then we have $k\in C_0$ for $k\le\omega$. Moreover, note that the "$h$-sequence" has already stablized by step $\omega$: no matter how we continue the sequence of $f$s, we'll always have $$h_\alpha^0(x)=x$$ for $\alpha\ge\omega$. In your notation, this means $N_0=\omega$.
But this doesn't really do anything to constrain the sequence of $f$s going forward: there is still "lots of room" between $f_\omega$ and $g$. Basically, $\lim_{\alpha\rightarrow\infty,\alpha\in C_0}h_\alpha^0$ doesn't really capture much information.
(Also, your last sentence is incorrect: we could have $C_0$ be everything!)

