$\newcommand\ep\epsilon $In the clever answer by Fedor Petrov, it was shown that 
\begin{equation*}
	Q:=P\Big(\frac{X_1^2+\dots+X_k^2}{X_1^2}<C\Big)\le C/k, \tag{1}\label{1}
\end{equation*}
where $C:=1/\ep^2>0$ and the $X_i$'s are any iid random variables. 

Let us show that for the standard normal $X_i$'s as in the OP, the upper bound $C/k$ in \eqref{1} can be replaced by a bound decreasing exponentially in $k$. 

For $C\le1$, $Q=0$. So, without loss of generality (wlog) $C>1$. Also, wlog $k\ge2$. 
For 
\begin{equation*}
	c:=C-1>0, \tag{2}\label{2}
\end{equation*}
any $x\in [0,\sqrt{(k-1)/c}\,]$, and 
$$h:=\frac{k-1-cx^2}{4(k-1)},$$ 
we have
\begin{equation*}
	Q=P\Big(\sum_{i=2}^k X_i^2<cX_1^2\Big)\le Q_1+Q_2, \tag{3}\label{3}
\end{equation*}
where 
\begin{equation*}
	Q_1:=P(X_1^2\ge x^2)\le e^{-x^2/2}=:R_1 \tag{4}\label{4}
\end{equation*}
and 
\begin{equation*}
\begin{aligned}
	Q_2&:=P\Big(\sum_{i=2}^k X_i^2<cx^2\Big) \\ 
	&=P\Big(\sum_{i=2}^k(1-X_i^2)>k-1-cx^2\Big) \\ 
	&\le\exp\{-h(k-1-cx^2)+(k-1)\ln Ee^{h(1-X_1^2)}\} \\ 
	&=\exp\{-h(k-1-cx^2)+(k-1)(h-\tfrac12\,\ln(1+2h))\} \\ 
	&\le\exp\{-h(k-1-cx^2)+2(k-1)h^2\} \\ 
	&=\exp-\frac{(k-1-cx^2)^2}{8(k-1)}=:R_2. 
\end{aligned}
\tag{5}\label{5}
\end{equation*}
Choosing now $x$ to be the positive root of the equation $R_1=R_2$, from \eqref{3}, \eqref{4}, and \eqref{5} we get 
\begin{equation}
	Q\le 2\exp-\frac{k-1}{2(1+\sqrt C)^2}, \tag{6}\label{6}
\end{equation}
which is the promised bound, decreasing exponentially in $k$. 

One may also note that typically $k$ and $C$ will be large, and then the exponent in the bound in \eqref{6} will be about $-k/(2C)$ -- compare this with the bound in \eqref{1}.