Here is a proof that such a polynomial does not exist assuming that every admissable $n$-tuple occurs infinitely often in the sequence of primes.
To see this let $(0, a_1, \dots, a_{n-1})$ be an admissable $n$-tuple that occurs infinitely often. Suppose $P \in \mathbb{Z}[x_1, \dots, x_n]$ is such that the function $f(i):=P(p_i,p_{i+1},\dots p_{i+n-1})$ is bounded. Replacing $x_i$ by $x_1+a_{i-1}$ for all $i \in \{2, \dots, n\}$ we obtain a polynomial $Q \in \mathbb{Z}[x_1]$. Since the tuple $(0, a_1, \dots, a_{n-1})$ occurs infinitely often, we have $Q(p_i)=f(i)$ for infinitely many $i$. Since $f(i)$ is bounded, $Q$ must be a constant polynomial, and thus $P$ is also a constant polynomial.