$\newcommand\Z{\mathbf Z}\newcommand\ep\varepsilon\newcommand\tP{\tilde P}\newcommand\de\delta\newcommand\R{\mathbb R}$Note that the random point $\Z/|\Z|$ is uniformly distributed on the sphere $S^{n-1}$, where $\Z:=(Z_1,\dots,Z_n)$, $|\Z|:=\sqrt{Z_1^2+\dots+Z_n^2}$, and $Z,Z_1,\dots,Z_n$ are iid standard normal random variables. So, the probability in question is \begin{equation} P:=P(Z_i^2\le Z_1^2-a|\Z|^2\ \forall i\ge2). \end{equation} Let \begin{equation} a=\frac{c\ln n}n, \end{equation} where $c$ is a nonnegative real constant; we shall see that this is the right choice for $a$.
Using your favorite exponential inequality, we see that eventually (that is, for all large enough $n$) \begin{equation} P(|\,|\Z|^2-n|>\sqrt n\,\ln n)\le\ep_n:=\exp\Big\{-\frac{\ln^2n}3\Big\}. \end{equation} So, \begin{equation} P=\tP+O(\ep_n), \end{equation} \begin{equation} \tP:=P(Z_i^2\le Z_1^2-b\ \forall i\ge2), \end{equation} where \begin{equation} b=na(1+\de_n)=(1+\de_n)c\ln n =c\ln n+o(1)\quad\text{and}\quad|\de_n|\le\frac{\sqrt n\,\ln n}n=o(1/\ln n). \end{equation}
Next, conditioning on $Z_1$, we have \begin{equation} \tP=\int_\R dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1}, \end{equation} where $f$ is the standard normal pdf, $G(t):=P(Z>t)$, and $t_+:=\max(0,t)$. Consider here the substitution \begin{equation} e^{-u}=1-2G(\sqrt{(x^2-b)_+}), \end{equation} with $e^{-u}:=0$ if $x^2-b\le0$; note that $u\ge0$. For each real $h>0$, the integral of $f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1}=f(x)e^{-(n-1)u}$ over the set of $x>0$ such that $u>h$ is less than $e^{-(n-1)h}$. So, we can choose $h=h_n$ so that \begin{equation} h\downarrow0,\quad nh\to\infty, \end{equation} and \begin{equation} \int_{x>0\colon\, u>h} dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1} \le\ep_n \end{equation} eventually.
On the other hand, if $u\le h$ and $x>0$, then $u=o(1)$ and hence \begin{equation} \sqrt{(x^2-b)_+}=\sqrt{x^2-b}=z:=G^{-1}\Big(\frac{1-e^{-u}}2\Big)\to\infty, \end{equation} \begin{equation} x=\sqrt{b+z^2}, \end{equation} \begin{equation} dx=-\frac{du\, z}{\sqrt{b+z^2}}\frac1{f(z)}\frac{e^{-u}}2. \end{equation} Next, $G(z)=\frac{1-e^{-u}}2\sim u/2$ and $G(z)=e^{-z^2/(2+o(1))}$, whence \begin{equation} z\sim\sqrt{2\ln\frac1u}, \end{equation} \begin{equation} f(z)\sim G(z)z\sim\frac u2\,\sqrt{2\ln\frac1u}, \end{equation} \begin{equation} dx\sim-\frac{du\,\sqrt{2\ln\frac1u}}{\sqrt{b+2\ln\frac1u}}\frac1{\frac u2\,\sqrt{2\ln\frac1u}}\frac{e^{-u}}2 =-\frac{du}{\sqrt{b+2\ln\frac1u}}\frac{e^{-u}}u. \end{equation} Further, \begin{equation} f(x)=e^{-b/2}f(z)\sim e^{-b/2}\frac u2\,\sqrt{2\ln\frac1u}. \end{equation} So, \begin{align*} & \int_{x>0\colon\, u\le h} dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1} \\ &\sim\int_0^h e^{-b/2}\frac u2\,\sqrt{2\ln\frac1u}\,\frac{du}{\sqrt{b+2\ln\frac1u}}\frac{e^{-u}}u e^{-(n-1)u} \\ &=\frac{e^{-b/2}}2\,\int_0^h du\,\frac{\sqrt{2\ln\frac1u}}{\sqrt{b+2\ln\frac1u}} e^{-nu} \\ &=\frac{e^{-b/2}}{2n}\,\int_0^{nh} dy\, e^{-y}\,\frac1{\sqrt{1+\frac b{2\ln\frac ny}}} \\ &=\frac{e^{-(c\ln n+o(1))/2}}{2n}\,\int_0^{nh} dy\, e^{-y}\,\frac1{\sqrt{1+\frac{c\ln n+o(1)}{2\ln\frac ny}}} \\ &\sim \frac{n^{-1-c/2}}2\,\frac1{\sqrt{1+c/2}} \end{align*} by dominated convergence.
Collecting all the pieces and noting that the integrand is even in $x$, we conclude that for $a=\frac{c\ln n}n$ \begin{equation} P\sim \frac{n^{-1-c/2}}{\sqrt{1+c/2}} \end{equation} uniformly over all nonnegative $c$ in any bounded interval.