A sequence $$n_{i+1} = f_m(n_i)$$ with $f_m$ a polynomial of degree $m$. With this also $n_i$ is a polynomial:
$$n_i=\sum_{j=0}^{m\cdot i} a_jx^j$$
If $n_1$ has polynomial degree $m$ then the degree of $n_i$ is $i$-times higher.
Given $$n_0= c \text{ , (constant)} >=0 \in \mathbb{R}$$ and a sequence value $v$. Is there any way to compute index $i$ with $$n_i=v$$ ?
Or is the only way to compute all sequence elements until $n_i$ equal with $v$?
(Especialy interested for $f$ a Chebyshev polynomial (first kind))
In general, no, esp. if the constant term is of the polynomial is zero. Consider the example where $m = 2$ and the polynomial $p(x)$ is $p(x) = x^2 - x$. For $n_0 = 1$, you get $n_1 = n_2 = \ldots = 0$. So if I tell you that $n_i = 0$ and ask you "What's $i$?" the answer is "1, or 2, or 3, or ...", i.e., the question doesn't even have a well-defined answer.
More generally, if $c$ happens to be a fixed point of the polynomial $p$, then the question has no answer; the same goes if $c$ is in the preimage of a fixed point, or the preimage of the preimage of a fixed point, etc.
Since every polynomial except for $p(x) = x + k$, ($k \ne 0$) has fixed points, the problem in general has no solution.
In the very special case of $p(x) = x + k$, if you know that $n_i = v$, then you know that $c + k \cdot i = v$, and you can compute $i$ via $i = \frac{v-c}{k}$.
Not very interesting, of course, but at least it's a case where the question has an answer.