g.c.d. and Euler's totient function 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.  
 A: 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.
A: [I assume that by "$a_i$ coprime" you mean that the $a_i$ have no common divisor, and not that they are pairwise coprime. That would make things tricky.]
Given a collection of Sophie Germain primes ($p_i$ such that $2p_i+1$ is a prime), we can construct families where the gcd grows exponentially in $n$, with $n$ the sum of the larger primes in each pair. 
First, Fermat's little theorem tells us that $2p+1$ divides $x^{2p+1} - x$ for any integer $x$. Take $x^{2p+1} + (p-1)x$ if you want to use only natural numbers.
Then $\phi(2p+1) = 2p$ divides $\phi(x^{2p+1} - x)$ for all $x$. 
From here, we let $P(x) = \Pi(x^{2p_i+1} - x)$ for some collection of Sophie Germain primes $p_i$. Then $\Pi p_i$ divides the gcd, and the degree is $\Sigma(2p_i + 1)$
