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

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. Thus $Q_a=p(a)$ where $p$ is a non-constant polynomial only depending on $P$. Since $P$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible.

As Terry Tao notes in the comments below:

One can make this argument unconditional by noting that $O(\log^{n−1−o(1)}X)$ of the $O(\log^{n−1}X)$ admissible tuples $a=(0,a_1, \dots ,a_{n−1})$ with $a_1, \dots, a_{n−1}=O(\log X)$ will be associated to consecutive primes $p,p+a_1, \dots,p+a_{n−1}$ for some $p∼X$ (because $∼X/\log X$ primes will generate a tuple by Markov's inequality and each tuple is associated to $O(X/\log^{n−o(1)}X)$ primes by e.g. Selberg sieve). On the other hand, a polynomial constraint on these tuples would instead force at most $O(\log^{n−2}X)$ of these tuples to be admissible (Schwartz-Zippel lemma).

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

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. Thus $Q_a=p(a)$ where $p$ is a non-constant polynomial only depending on $P$. Since $f$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible.

As Terry Tao notes in the comments below:

One can make this argument unconditional by noting that $O(\log^{n−1−o(1)}X)$ of the $O(\log^{n−1}X)$ admissible tuples $a=(0,a_1, \dots ,a_{n−1})$ with $a_1, \dots, a_{n−1}=O(\log X)$ will be associated to consecutive primes $p,p+a_1, \dots,p+a_{n−1}$ for some $p∼X$ (because $∼X/\log X$ primes will generate a tuple by Markov's inequality and each tuple is associated to $O(X/\log^{n−o(1)}X)$ primes by e.g. Selberg sieve). On the other hand, a polynomial constraint on these tuples would instead force at most $O(\log^{n−2}X)$ of these tuples to be admissible (Schwartz-Zippel lemma).

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

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. Thus $Q_a=p(a)$ where $p$ is a non-constant polynomial only depending on $P$. Since $P$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible.

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

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. Thus $Q_a=p(a)$ where $p$ is a non-constant polynomial only depending on $P$. Since $P$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible.

As Terry Tao notes in the comments below:

One can make this argument unconditional by noting that $O(\log^{n−1−o(1)}X)$ of the $O(\log^{n−1}X)$ admissible tuples $a=(0,a_1, \dots ,a_{n−1})$ with $a_1, \dots, a_{n−1}=O(\log X)$ will be associated to consecutive primes $p,p+a_1, \dots,p+a_{n−1}$ for some $p∼X$ (because $∼X/\log X$ primes will generate a tuple by Markov's inequality and each tuple is associated to $O(X/\log^{n−o(1)}X)$ primes by e.g. Selberg sieve). On the other hand, a polynomial constraint on these tuples would instead force at most $O(\log^{n−2}X)$ of these tuples to be admissible (Schwartz-Zippel lemma).

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

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. Thus $Q_a=p(a)$ where $p$ is a polynomial only depending on $P$. Since $P$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible.

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

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. Thus $Q_a=p(a)$ where $p$ is a non-constant polynomial only depending on $P$. Since $P$ takes only finitely many values, $p(a)$ only takes on finitely many values over all admissible $n$-tuples $a$. However, this is impossible.