It was already pointed out in the comments that determining for which $n$ one can find a non-constant $p_n$ is an open problem. I thought I'd give a bit of context and my understanding on what is known so far. The problem as stated has a negative answer because when $n+1$ is prime, $p_n$ must be constant.

The sums $\sum_{i=0}^n \varepsilon_i \binom{n}{i}$ are the leading coefficients of the polynomials we get from Lagrange interpolation on points $(i,\alpha(i))$ where $0\le i\le n$ and $\alpha(i)\in \lbrace 0,1\rbrace$. So the question is equivalent to

> Is there a polynomial that sends $\lbrace 0,1,\dots,n\rbrace$ to $\lbrace 0,1\rbrace$, that is not constant but has degree $\le n-1$?

Let us denote the number of such polynomials by $\mathcal B(n)$. Some examples are given by $\varepsilon_i=(-1)^i$ when $n$ is even and $\varepsilon_i=-\varepsilon_{n-i}$ for odd $n$. This implies that $\mathcal B(n)\geq 2$ when $n$ is even and $\mathcal B(n)\geq 2^{\frac{n+1}{2}}$ when $n$ is odd. Finding non-trivial solutions to the problem implies improving on these lower bounds.

Here is a simple argument, that when $p$ is an odd prime $\mathcal B(p-1)=2$, so there are no non-trivial solutions. This is because $\binom{p-1}{i}\equiv (-1)^i\pmod{p}$ so the only way for the sum to be divisible by $p$ is if the sequence $(-1)^i\varepsilon_i$ is constant. This includes the examples $n=16,18$ that you confirmed with a computer search. However there are even values of $n$ for which $k(n)\geq 3$. The first example is
$$\binom{8}{0}-\binom{8}{1}-\binom{8}{2}+\binom{8}{3}+\binom{8}{4}-\binom{8}{5}-\binom{8}{6}-\binom{8}{7}+\binom{8}{8}=0$$
and the next one is the one given by Darij in the [comments][1] for $n=14$. The even values of $n$ for which $\mathcal B(n)\geq 3$ and $n\le 128$ were found in J. von zur Gathen and J. Roche, ["Polynomials with two values"][2], Combinatorica 17, no. 3 (1997), 345–362. The sequence is $\lbrace 24,34,48,54\rbrace$ and numbers $2\pmod{6}$.

Your question is really about proving that $\mathcal B(n)=2$ for infinitely many $n$, and it is an open to determine such $n$ besides the values found in the von Zur Gathen-Roche paper. As I mentioned above it is equivalent to proving for such $n$ that the minimum degree of a non-constant polynomial sending $\lbrace 0,1,\dots,n\rbrace\to \lbrace 0,1\rbrace$ is $n$. The best results known so far are that the degree is $n-o(n)$, where the $o(n)$ comes from the gaps in consecutive primes (so one can take $O(n^{.525})$), but conjecturally this can be improved to $n-O(1)$. 

One thing that is surprising is the following threshold phenomenon. If we look at non-constant polynomials sending $\lbrace 0,1,\dots,n\rbrace\to \lbrace 0,1,\dots,n\rbrace$, the minimum degree is $1$ (for instance $f(x)=x$), but as soon as we look at $\lbrace 0,1,\dots,n\rbrace\to \lbrace 0,1,\dots,n-1\rbrace$ the degree is at least $n-o(n)$. The current methods don't seem to make use of the fact that in the boolean case the range is simply $\lbrace 0,1\rbrace$, as they give the same bound for larger ranges. A recent paper on the topic is ["On the Degree of Univariate Polynomials Over the Integers"][3] by G. Cohen, A. Shpilka and A. Tal.


  [1]: https://mathoverflow.net/questions/74191/what-is-the-degree-of-a-symmetric-boolean-function#comment188980_74191
  [2]: https://doi.org/10.1007/BF01215917 "zbMATH review at https://zbmath.org/0907.68134"
  [3]: https://doi.org/10.1007/s00493-015-2987-0 "Conference version available at doi:10.1145/2090236.2090269. zbMATH reviews at https://zbmath.org/1399.11080|1348.11024"