Suppose that $x_1,\ldots,x_n$ are positive real numbers such that $$ x_1^2+\cdots +x_n^2 < \epsilon. $$ Can we bound the quantity $$ 2^{-n}\sum_{b_1,\ldots,b_n\in\{\pm1\}}e^{\left(\sum_i b_ix_i\right)^2}? $$ `Bound' means the following: For all sufficiently small $\epsilon$, find a constant $C_{\epsilon}$ such that $$ 2^{-n}\sum_{b_1,\ldots,b_n\in\{\pm1\}}e^{\left(\sum_i b_ix_i\right)^2}\leq C_{\epsilon}. $$
Partial results: When $n=2$ and $\epsilon=1/2$, I can prove the following inequality: $$ \frac{e^{(x+y)^2}+e^{(x-y)^2}}{2}\leq \left(1-2\epsilon\right)^{-1/2}. $$ I believe an inequality of the following form should hold: $$ \frac{\sum_{\pm}e^{(x_1\pm x_2\pm \cdots x_n)^2}}{2^n}\leq e^{\sum_i x_i^2}, $$ although I've run into some issues proving such a bound.
Notice that each term in the sum is bounded by
$$ e^{(\sum_{i}x_i)^2} = e^{\sum_{i} x_i^2 + \sum_{i \neq j} x_i x_j} \leq e^{\epsilon + \sum_{i \neq j} x_i x_j} $$ Then observe that the vector formed by the $x_i$ lies inside a ball of radius $\epsilon$, so each $x_i < \sqrt{\epsilon}$ and we can bound the sum
$$ \sum_{i \neq j} x_i x_j < \sum_{i \neq j} \epsilon =n(n-1)\epsilon $$ hence the sum above is bounded by
$$ e^{(\sum_{i}x_i)^2} < e^{\epsilon(1 + n(n-1))} $$