EDITED: As pointed out by Anthony and John, my 2am solution was anything but. In summary, the conjecture is TRUE for $C$ smaller than approximately 0.6880137 and false for larger $C$.

The exact value is
$$ S(C,n) = \sum_{k=0}^n 2^{-n}\binom{n}{k} \exp\bigl(Cn^{-3}(n-2k)^4\bigr)$$
since the product of the first two terms is the probability that $\sum_i X_i=n-2k$.

Consider the term $k=(\tfrac12+\beta)n$, where $-\frac12\le\beta\le\frac12$. Stirling's approximation for constant $\beta\ne\pm\frac12$ is
$$2^{-n}\binom nk \approx \bigl(2\pi(\tfrac12-\beta)(\tfrac12+\beta)n\bigr)^{-1/2}
\bigl(2 (\tfrac12-\beta)^{\tfrac12-\beta} (\tfrac12+\beta)^{\tfrac12+\beta}\bigr)^{-n}$$

Therefore the value of the term is close to
$$\bigl(2\pi(\tfrac12-\beta)(\tfrac12+\beta)n\bigr)^{-1/2} \exp\bigl(f(\beta)n\bigr),$$
where
$$ f(\beta) = -(\tfrac12-\beta)\ln(\tfrac12-\beta)
- (\tfrac12+\beta)\ln(\tfrac12+\beta)
- \ln 2 + 16C\beta^4. $$
(Note that there is an order $1/n$ term from Stirling's formula missing. Since it doesn't depend on $C$ and the value of the sum is 1 for $C=0$, the effect of the missing term is easily corrected at the end.)

The function $f(\beta)$ has a local maximum of value 0 at $\beta=0$ for every $C$.
There is a constant $C_0\approx 0.6880137$ (note: less than $\ln 2$) so that for $C\lt C_0$ we have $f(\beta)\lt 0$ for all $\beta\ne 0$. For $C\gt C_0$ there are places where $f(\beta)\gt 0$ and the sum will be exponential.

Therefore, when $C\lt C_0$ the sum is dominated by terms near $\beta=0$. Around there,
$$f(\beta) = -2\beta^2 + (16C-\tfrac43)\beta^4 + O(\beta^6),$$
so the term is close to
$$(\pi n/2)^{-1/2} e^{-2\beta^2 n}\bigl(1+(16C-\tfrac43)\beta^4n\bigr) $$
with terms for $\beta=O(n^{-1/2})$ mattering. Approximating the sum by an integral, I get
$$ S(C,N) = 1 + 3C/n + O(n^{-2}),$$
for $C\lt C_0$.

ADDED: The exact value of $C_0$ is
$$\frac{1}{64\beta_0^3}\ln\Bigl(\frac{1+2\beta_0}{1-2\beta_0}\Bigr),$$
where $\beta_0$ is the zero near 0.4953 of
$$\bigl(-\tfrac12+\tfrac34\beta\bigr)\ln(1-2\beta) +
\bigl(-\tfrac12-\tfrac34\beta\bigr)\ln(1+2\beta).$$
To 50 digits: $C_0 = 0.68880137394879099153980106986892429419403651844842$.