Pick two unit vectors. Rotate your coordinate system so that one of them points at $(1,0,\ldots,0)$. You want the probability that $x \cdot (1,0,\ldots, 0) < T$. It will suffice to find the proportion of the surface area of the $n$-sphere not within an angle of $\cos^{-1}(T)$ from the North pole.
There are some nice formulas for the area of spherical caps in this paper by S. Li. In particular, the area of a cap with angle $\phi \leq \pi/2$ is $$A_n^{cap} = \frac{2\pi^{\frac{n-1}{2}}}{\Gamma(\frac{n-1}{2})} \int_0^\phi \sin^{n-2}(\theta)d\theta.$$
On the other hand, the surface area of the $n$-sphere is known to be $$\frac{2\pi^{\frac{n+1}{2}}}{\Gamma(\frac{n+1}{2})}$$
Taking the quotient of these two, you'll get the proportion of there sphere within an angle of $\phi$ from the north pole. Now subtract that from $1$ for your answer.
We shall follow a geometric approach.
Let $S\subset\Bbb R^n$ be a sphere of radius $r=\sqrt{n}$ centered at the origin $0$ of $\Bbb R^n$. If $v=0$ then $|\langle a, v \rangle|\ge 0$ for any $a\in S^{n-1}$. If $v\ne 0$ then it is easy to check that a set $\{a\in S:|\langle a, v \rangle| \geq 1/2 \|v\|_2\}$ consists of two spherical caps, cut from $S$ by hyperplanes orthogonal to $v$ and placed at a distance $1/2$ from the origin. It follows that each cap is spanned by an angle $2\psi$, where $\cos\psi=\tfrac 1{2r}=\tfrac 1{2\sqrt{n}}$.
When $n=1$ then $S$ consists of two points covered by the caps. When $n=2$ then $S$ is a circle, $\cos\psi=\tfrac 1{2\sqrt{2}}<\tfrac 1{1\sqrt{2}}=\cos\tfrac{\pi}4$, so $\psi>\tfrac{\pi}4$ and the caps cut more than half of the circle length.
So from now we assume $n\ge 3$.
We have that ($n-1$) dimensional area of $S$ equals $$A_n=r^{n-1}\frac{2\pi^{n/2}}{\Gamma\left(\tfrac n2\right)}.$$
According to [Li], an area $C_n$ of the spherical cap equals $$r^{n-1}\frac{2\pi^{(n-1)/2}}{\Gamma\left(\tfrac {n-1}2\right)}\int_0^\phi\sin^{n-2} \theta d\theta,$$ where, as I understood, $2\phi$ is the angle spanned by the cap.
Thus we have to show that the area remaining after cutting the caps is at most half of the sphere area, that is
$\frac{2\pi^{(n-1)/2}}{\Gamma\left(\tfrac {n-1}2\right)}\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \frac 14 \frac{2\pi^{n/2}}{\Gamma\left(\tfrac n2\right)}$
$\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}.$
Since $n\ge 3$,
$$\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \int_\psi^{\pi/2}\sin \theta d\theta=-\cos\frac \pi{2}+\cos \psi=\tfrac 1{2r}=\tfrac 1{2\sqrt{n}}.$$
Now we estimate the right-hand side of the inequality.
By Legendre’s formula, $$\Gamma\left(\tfrac {n-1}2\right)\Gamma\left(\tfrac n2\right)=\frac{\sqrt{\pi}}{2^{n-2}}
\Gamma(n-1)=\frac{(n-2)!\sqrt{\pi}}{2^{n-2}}.$$
1)) If $n$ is even then $\Gamma\left(\tfrac n2\right)=\left(\tfrac n2-1\right)!$, so
$$\frac{\Gamma\left(\tfrac {n-1}2\right)}{\Gamma\left(\tfrac n2\right)}= \frac{(n-2)!\sqrt{\pi}}{2^{n-2}\left(\tfrac n2-1\right)!^2}=\frac{\sqrt{\pi}}{2^{n-2}}{n-2\choose \tfrac {n-2}2}.$$
Robbins’ bounds imply that for any positive integer $m$
$$\frac {4^{m}}{\sqrt{\pi m}}\exp\left(-\frac {1}{8m-1}\right)<{2m\choose m}<\frac {4^{m}}{\sqrt{\pi m}}\exp\left(-\frac {1}{8m+1}\right).$$
So
$$\frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}=\frac{\pi}{2^n}{n-2\choose \tfrac {n-2}2}>
\frac{\pi}{2^n} \frac {2^{n-2}}{\sqrt{\pi (n-2)/2}}\exp\left(-\frac {1}{4n-9}\right)=$$
$$\sqrt{\frac{\pi}{8n-16}}\exp\left(-\frac {1}{4n-9}\right)>\sqrt{ \frac{\pi}{8n}}\exp\left(-\frac {1}{7}\right)
>\frac{0.54}{\sqrt{n}}>\frac 1{2\sqrt{n}}.$$
2)) If $n$ is odd then $\Gamma\left(\tfrac {n-1}2\right)= \left(\tfrac {n-3}2\right)!$, so
$$\frac{\Gamma\left(\tfrac {n-1}2\right)}{\Gamma\left(\tfrac n2\right)}=
\frac{2^{n-2}\left(\tfrac {n-3}2\right)!^2}{(n-2)!\sqrt{\pi}}=\frac{2^{n-2}}{(n-2)\sqrt{\pi}}{n-3\choose \tfrac {n-3}2}^{-1}.$$
If $n=3$ then the latter expression equals $\tfrac {2}{\sqrt\pi}>\tfrac 1{2\sqrt{n}}$. If $n>3$ then Robbins’ bounds imply that
$$\frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}=
\frac{2^{n-4}}{n-2}{n-3\choose \tfrac {n-3}2}^{-1}>\frac{2^{n-4}}{n-2}\cdot \frac {\sqrt{\pi (n-3)/2}}{2^{n-3}}\exp\left(\frac {1}{4n-11}\right)>\frac {\sqrt{\pi (n-3)/2}}{2(n-2)}> \frac 1{2\sqrt{n}}.$$
References
[Li] S. Li, Concise Formulas for the Area and Volume of a Hyperspherical Cap, Asian Journal of Mathematics and Statistics 4:1 (2011) 66–70.
Best Answer
As discussed, the question here relates to this question. Fix $n\in\mathbb Z_{>0}$. Let $X_1,X_2\sim\mathcal N(0,I_n)$ be independent standard normal, let $Z_i=\|X_i\|_2$ and $X^\circ_i=\frac{1}{Z_i}X_i$ for $i\in\{1,2\}$. We know the following distributions.
What we do not know, is the distribution of $X^\circ_1\cdot X^\circ_2$, in particular it is not Gaussian (remember that $|X^\circ_1\cdot X^\circ_2|\le 1$ almost surely, this is certainly not normally distributed). So, what do we know about $X^\circ_1\cdot X^\circ_2$?
Since we do not know the distribution of $V$ and $X^\circ_1\cdot X^\circ_2$, there is little hope to obtain probabilities of events, say $\mathbb P(|X^\circ_1\cdot X^\circ_2|\le\alpha)$. What we have understood, however, are the asymptotics. I mean, we have just said that we know that $\lim_{n\rightarrow\infty}\mathbb P(|V|\le\alpha)=\mathbb P(|W|\le\alpha)$, for all $\alpha\in\mathbb R_{\ge 0}$. At this point, we really have to pay attention.
The answer in this post addresses a random variable $Y\sim\mathcal N(0,1/n)$. How does this fit here?
This explains that we cannot (yet reliably) say that $X=u\cdot v$ has the law $\mathcal N(0,1/n)$, not even in an approximate sense. If you want to take a closer look at the finite size behavior of $X^\circ_1\cdot X^\circ_2$, I'd be happy to, but maybe in a new question. In the remainder of this answer I will focus on $Y\sim\mathcal N(0,1/n)$ and $\mathbb P(|Y|\le\alpha)$, since this was the actual question.
So, let $\alpha\in\mathbb R$. Recall from above that we have $\sqrt{n}Y\sim\mathcal N(0,1)$, so $\sqrt{n}Y$ and $W$ have the same law and we can write $\mathbb P(|Y|\le\alpha)=\mathbb P(|W|\le\sqrt{n}\alpha)$. The question is why we have $\mathbb P(|W|\le\sqrt{n}\alpha)=\sqrt{n}\alpha$, the answer is: we don't. We have $$\mathbb P(|W|\le\sqrt{n}\alpha)=\int_{-\sqrt{n}\alpha}^{\sqrt{n}\alpha}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\mathrm{d}x.$$ Some people would write $\mathbb P(|W|\le\sqrt{n}\alpha)=\Phi(\sqrt{n}\alpha)-\Phi(-\sqrt{n}\alpha)=1-2\Phi(-\sqrt{n}\alpha)$, where $\Phi$ denotes the cumulative distribution function of $\mathcal N(0,1)$. There's no better way to write this.
Now, let's turn to the claim. In fact, we can get somewhat decent approximations. Let's use the shorthand $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$. As explained above, we want to know how $\mathbb P(|W|\le\sqrt{n}\alpha)$ and $\sqrt{n}\alpha$ are related for $0\le\alpha\le 1/\sqrt{n}$. So, let's do ourselves a favor first and just consider $\mathbb P(|W|\le a)$ and $a$ for $0\le a\le 1$, which is equivalent (to give some background, we look at the probability to be within an interval of at most one standard deviation from the mean, that's how it was intended in the answer in the first place). Using that $\phi$ has its unique maximum at $0$, we obtain the bounds $e^{-1/2}\sqrt{\frac{2}{\pi}}a=\frac{2e^{-1/2}a}{\sqrt{2\pi}}\le 2a\phi(1)\le 2a\phi(a)\le\mathbb P(|W|\le a)\le 2a\phi(0)=\frac{2a}{\sqrt{2\pi}}=\sqrt{\frac{2}{\pi}}a$. What does this tell us? It tells us that, strictly speaking, we have $\mathbb P(|W|\le a)\le \sqrt{\frac{2}{\pi}}a<a$, so the assumption $\mathbb P(|W|\le a)=a$ is wrong. But, not so strictly speaking, we have $\mathbb P(|W|\le a)=c(a)a$, where $e^{-1/2}\sqrt{\frac{2}{\pi}}\le c(a)\le \sqrt{\frac{2}{\pi}}$ is a coefficient depending on $a$, somewhere between $0.4839$ and $0.7979$, so the error is not catastrophic using this approximation. Since the argument in the answer was not formalized (and doesn't depend on constants, only on orders), this approximation is reasonable.
As I mentioned before, if you want to understand the error made by approximating $X^\circ_1\cdot X^\circ_2$ with $Y$ (respectively $\frac{1}{\sqrt{n}}W$), then please post a new question, I'd be happy to look into it.