$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm om $\R^n$. Take any real $t$. Let $Z$ be a random vector in $\R^n$ such that (i) $Z$ is independent of $X$ and (ii) $Z\sim N(0,\Sigma_Y-\Sigma_X)$. Then $X+Z$ equals $Y$ in distribution.
So, it suffices to show that \begin{equation} P(\|X\|\le t)\le P(\|X+Z\|\le t). \end{equation}
But this follows because (i) $P(\|X+Z\|\le t)=Eg(Z)$ with $g(z):=P(\|X+z\|\le t)$ and (ii) $g$ is an even log-concave function and hence $g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$.