Here is a proof that such a polynomial does not exist assuming that every admissible [$n$-tuple][1] occurs infinitely often in the sequence of primes.  

  [1]: https://en.wikipedia.org/wiki/Prime_k-tuple

To see this let $a:=(0, a_1, \dots, a_{n-1})$ be an admissible $n$-tuple. 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 $Q_a \in \mathbb{Z}[x_1]$.  Assuming that the $n$-tuple $a$ occurs infinitely often, we have $Q_a(p_i)=f(i)$ for infinitely many $i$.  Since $f(i)$ is bounded, $Q_a$ is a constant.  Let $d$ be the degree of $P$.  It is easy to show that $d \geq 2$.  Since the coefficient of $x_1^{d-1}$ in $Q_a$ is zero, $L(a)=0$, where $L$ is a linear function depending only on $P$.  Assuming that every admissible $n$-tuple occurs infinitely often, we have that $L(a)=0$ for every $a$.  However, this is impossible (going to sleep now, but will add more details later if this is unclear).