I would like to find a closed form for this sum: $$f_n(x) = \frac{1}{2^n} \sum_{k=0}^{n} \binom{n}{k} e^{(n-2k)^2 x} = \frac{1}{2^n} \sum_{k=0}^{n} \binom{n}{k} y^{(n-2k)^2} \quad\text{for}\ y = e^x$$ as a function of $x \in \mathbb{R}$ for $n \in \mathbb{N}$. $f_n(x)$ is a polynomial in $y = e^x$ of degree $n^2$, i.e., $f_n = b_n y^{n^2} + b_{n-2} y^{(n-2)^2} + \cdots$.
Added background: This sum arises in a physical problem that reduces to evaluating the average of $\cos^n \theta$: $$f_n(-\sigma^2/2) = \overline{\cos^n \theta} = \int_{\Gamma} d\theta P(\theta) \cos^n \theta,$$ where $\Gamma$ is any interval of length $2\pi$ and $P(\theta)$ is the wrapped normal distribution with zero mean and variance $\sigma^2$: $$P(\theta) = \frac{1}{\sqrt{2\pi} \sigma} \sum_{j = -\infty}^{\infty} e^{-(\theta + 2\pi j)^2 / 2 \sigma^2} = \frac{1}{2\pi} \sum_{k = -\infty}^{\infty} e^{i k \theta - k^2 \sigma^2 /2}.$$
I would guess that the way to find a closed form (if it exists) is by using the Egorychev method (as described in these notes by Marko Riedel), but I'm not skillful enough to do this myself.
As a side note, this sum can be reformulated as the series: $$f_n(x) = \sum_{m=0}^{\infty} C(m,n) x^m, \quad C(m,n) = \frac{1}{m! 2^n} \sum_{k=0}^{n} \binom{n}{k} (n-2k)^{2m},$$ where the coefficients $C(m,n)$ are a bit similar to the Stirling numbers of the second kind, but still different.
Added: $C(m,n)$ is itself a polynomial of degree $m$ of variable $n$, which has the form: $$C(m,n) = \sum_{p=1}^{m} a_p n^p ,$$ where $a_m = (2m-1)!!/m!$ and $a_0$ is always zero. Formally, $C(m,n)$ can be expressed in terms of the generalized hypergeometric function: $$C(m,n) = \frac{n^{2m}}{m! 2^n} {}_{2m+1}\! F_{2m}\bigg(\underbrace{1 - \frac{n}{2}, \ldots, 1 - \frac{n}{2}}_{2m\ \mathrm{times}}, -n; \underbrace{-\frac{n}{2}, \ldots, -\frac{n}{2}}_{2m\ \mathrm{times}};-1 \bigg),$$ which is formally valid only for odd $n$ ($-n/2 \notin \mathbb{Z}$), but for even $n$ singularities in $\Gamma$ functions cancel out.