I think that you have found the lowest degree polynomial there is.
If $q=2$, then the result of your other recent question shows (among other things) that a polynomial of degree $\le n-1$ takes the value $1$ an even number of times on all of $\mathbf{Z}_2^n$ (=a cube of dimension $n$), but OR $=1$ exactly $2^n-1$ times.
=====================================
Added this to explain why allowing $q>2$ does not really change anything.
Let $S_n=\{0,1\}^n$, $q$ a prime (power), and $F_q=GF(q)$ the finite field of $q$ elements. Then we can view $S$ as a subset of $F_q^n$. An $n$-fold Lagrangian interpolation shows that any function $f:S\rightarrow F_q$ can be represented by a polynomial $g(x_1,x_2,\ldots,x_n)\in F_q[x_1,x_2,\ldots,x_n]$, i.e. $f(b_1,b_2,\ldots,b_n)=g(b_1,b_2,\ldots,b_n)$ for all $(b_1,b_2,\ldots,b_n)\in S.$
Let us consider the set of polynomials $g$ yielding the same function $f$. These form a coset of the ideal $I$ of polynomials vanishing on all of $S$.
Theorem. The ideal $I$ is generated by polynomials $x_i^2-x_i$, $i=1,2,\ldots,n$. Each coset of $I$ has a unique polynomial in the span of monomials of the set ${\mathcal B}=\{x_{i_1}x_{i_2}\cdots x_{i_k}\mid i_1 A polynomial $g$ in the $F_q$-span of ${\mathcal B}$ has the lowest possible degree in its coset $g+I$.
Proof. Obviously the polynomials $p_i=x_i^2-x_i$ all belong to $I$. Let $J$ be the ideal generated by $p_1,p_2,\ldots,p_n$, so $J\subseteq I$. Let $g$ be any polynomial. If any of the terms $a x_{i_1}^{e_1}x_{i_2}^{e_2}\cdots x_{i_k}^{e_k}$ of $g$ has a factor with exponent $e_i>1$ we can replace that factor with $x_i$. This is because the polynomial functions $x_i$ and $x_i^{e_i}$ agree on all of $S$, whenever $e_i>1$. Because $x_i^{e_i}\equiv x_i \pmod{p_i}$, we are moving within the coset $g+J$. Doing this for all the terms of $g$ shows that $g$ can be replaced with a polynomial $g'$ such that
- $\deg g'\le \deg g$,
- $g'\equiv g \pmod J$ (or, equivalently $g'+J=g+J$),
- $g'$ is in the span of ${\mathcal B}$.
But $|{\mathcal B}|=2^n=|S|$, and the dimension of the space $V$ of functions $S\rightarrow F_q$ is also $|S|=2^n$. Therefore $ \dim F_q[x_1,x_2,\ldots,x_n]/I=|S|=\dim F_q[x_1,x_2,\ldots,x_n]/J, $ so it is impossible that $J$ would be a proper subset of $I$. Therefore $I=J$ and the above process of replacing $x_i^{e_i}$ with $x_i$ in all the terms of $g$ for all $i$ leads to the lowest degree polynomial $g'$ in the coset $g+I=g+J$. Q.E.D.
The claim that the lowest degree polynomial representing $OR(x_1,x_2,\ldots,x_n)$ is the one you gave follows from this. This is because your polynomial is in the span of the monomials of ${\mathcal B}$.