There is this really nice paper by J.P.Serre on the congruence subgroup property for $SL_2$ for $S$-arithmetic groups (https://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result.
Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient function. For each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers
$$N(a,b)= g.c.d. \{ \phi (ax+b): x=1,2,3,\cdots \}.$$ Now $N(a,b)$ seemingly depends on $a,b$ but it does not much: $N(a,b)$ divides $8$.
The proof of this uses Dirichlet's theorem on infinitude of primes.
If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$.
My question is : if I replace the linear polynomial $ax+b$ by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d.
$$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial?
The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.
[Edit] I should have added the link https://arxiv.org/abs/math/0409377.
[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.
Best Answer
I have made some computations which seem to corroborate the OP's conjecture, namely that for any $n$ there exists a $N$, such that for every polynomial $P$ of degree $n$, with positive integral coefficients and content 1, the quantity $$g(P):= g.c.d(\phi(P(x)),x \geq 1)$$ divides $N$.
For $n=1$, as the OP says, one can take $N=8$ as proved by Serre.
For $n=2$, it seems that one can take $N=2^4 3^2 = 144$. It seems even more that one cannot do better, because for $P(x)=16x^2+32x+17$, I get experimentally $g(P)=16$ (this must not be hard to prove but I haven't tried), and for $P(x)=27 x^2 + 9x+1$, I get $g(P)=18$. So $144 | N$. On the other hand I have need been able to find any $P$ such that $g(P)$ was not a divisor of $144$.
For $n=3$ or $n=4$, I have failed to find any $P$ with $g(P)\geq 2$. This suggests $N=2$ in these cases.