$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm on $\R^n$. 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)\ge P(\|X+Z\|\le t). \tag{1}
\end{equation*}

Note that 
\begin{equation*}
	P(\|X+Z\|\le t)=Eg(Z), \tag{2}
\end{equation*}
where 
\begin{equation*}
	g(z):=P(\|X+z\|\le t)=\int_{\R^n}dx\, f(x)1(\|x+z\|\le t)
\end{equation*}
and $f$ is the pdf of $X$. The functions $f$ and $x\mapsto1(\|x+z\|\le t)$ are log concave, and hence the function $x\mapsto f(x)1(\|x+z\|\le t)$ is log concave. 
So, by the [Prékopa–Leindler theorem][1], $g$ is a log-concave function. Also, the function $g$ is even. So, 
$g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$, and hence (1) follows from (2). 
  
[1]: https://en.wikipedia.org/wiki/Pr%C3%A9kopa%E2%80%93Leindler_inequality#Applications_in_probability_and_statistics