As Paata suggests, we write 
$$
\mathbb{E} e^{tZ^{2}} = 2t \int_{0}^{\infty}\lambda e^{t\lambda^{2}}P(Z>\lambda)d\lambda. \quad\quad\quad (\heartsuit)
$$
Next, for any vector $(\delta_1,\ldots,\delta_n)\in \{-1,1\}^n$ we have $$P\left(\sum \delta_i X_i>\lambda\right)\leqslant P(Z>\lambda)\leqslant \sum_{\varepsilon_i=\pm 1,i=1,\ldots,n} P\left(\sum \varepsilon_i X_i>\lambda\right),$$
thus the integral $(\heartsuit)$ converges if and only if each integral
$$
 \int_{0}^{\infty}\lambda e^{t\lambda^{2}}P\left(\sum \delta_i X_i>\lambda\right)d\lambda
$$
converges. Since each $\sum \delta_iX_i$ is a 1-dimensional Gaussian with variance which I denote $\sigma^2(\delta_1,\ldots,\delta_n)$, you may take $t<1/(2\sigma^2(\delta_1,\ldots,\delta_n))$ and can not take $t\geqslant 1/(2\sigma^2(\delta_1,\ldots,\delta_n))$. (Possibly for specific $\delta_i$'s you can take $t=2\sigma^2(\delta_1,\ldots,\delta_n)$, but then for $-\delta_i$'s you can't.)