Note that $(1+x)^n$ itself does not have Gaussian coefficients when you go that far from the central term: by Stirling, for fixed $\rho \in (0,1)$ the $x^{\rho n}$ coefficient $\bigl( {n \atop \rho n} \bigr)$ is roughly proportional to $\exp H(\rho)$ where $H(\rho) = -\rho \log \rho - (1-\rho) \log(1-\rho)$. Your graph looks like it could be something like this multiplied by $\cos cx$ for some $c \approx 1/6$. The stationary-phase technique for finding the asymptotic behavior of power-series coefficients often gives rise to expressions such as this, though $H(r)$ might be a sum of more complicated terms than just $-\rho \log \rho$ and $-(1-\rho) \log(1-\rho)$.

Note that $(1+x)^n$ itself does not have Gaussian coefficients when you go that far from the central term: by Stirling, for fixed $\rho \in (0,1)$ the $x^{\rho n}$ coefficient $\bigl( {n \atop \rho n} \bigr)$ is roughly proportional to $\exp nH(\rho)$ where $H(\rho) = -\rho \log \rho - (1-\rho) \log(1-\rho)$. Your graph looks like it could be something like this multiplied by $\cos cx$ for some $c \approx 1/6$. The stationary-phase technique for finding the asymptotic behavior of power-series coefficients often gives rise to expressions such as this, though $H(r)$ might be a sum of more complicated terms than just $-\rho \log \rho$ and $-(1-\rho) \log(1-\rho)$.