Is it possible for some numbers $a, b, c \in \mathbb{R}$, $abc \neq 0$ that for every $n > 3$, there exists a polynomial $P_n(x) = x^n + a_{n-1}x + ... + ax^2 + bx + c$ such that all its roots are integers?
I have concluded with Vieta's formula that all coefficients of every polynomial $P_n$ must be integers.
Nope.
Let's say the polynomial has $N$ integer roots $\lambda_1,\ldots,\lambda_N \in \{ -1, 1 \}$ and $M$ integer roots $\lambda_{N+1},\ldots,\lambda_{N+M} \notin \{ -1, 1 \}$. Upto a sign, we have
$$\prod_{k=1}^M \lambda_{k+N} = \pm \prod_{k=1}^{N+M} \lambda_k = \pm c$$
So $M$ is bounded from above by $\Omega(c)$, the number of prime factors in $n$ (see wiki entry of prime omega function for definition/detail).
By Vieta's formulas, $N$ is also bounded.
$$N = \sum_{k=1}^N \frac{1}{\lambda_k^2} \le \sum_{k=1}^{N+M}\frac{1}{\lambda_k^2} = \frac{b^2 - 2ac}{c^2}$$
In order for such a polynomial to have all integer roots, its degree need to satisfy $$n = N + M \le \frac{b^2-2ac}{c^2} + \Omega(c)$$