Each coordinate $x_k$ is distributed like $R$ times a random variable with PDF $f_N$, where
$$
f_N(s)=\frac{\Gamma(\tfrac{N}2)}{\sqrt{\pi}\,\Gamma(\tfrac{N-1}2)}\cdot(1-s^2)^{\frac{N-3}2}\cdot[-1\leqslant s\leqslant1].
$$
The simplest approach to this result might be to remember that a standard normal vector $z=(z_k)$ is such that $z/\|z\|$ is uniformly distributed on the unit sphere. Hence, each $x_k$ is distributed like $R\cdot\xi$ where $\xi=z_1/\|z\|$.
Now, $\|z\|^2=z_1^2+r^2$, where $r\gt0$ is independent of $z_1$ with density proportional to $r^{N-1}\mathrm e^{-r^2/2}$. Forgetting some irrelevant normalizing constants, this yields that, for every bounded measurable function $u$,
$$
\mathrm E(u(\xi))\propto\iint u\left(\frac{w}{\sqrt{w^2+r^2}}\right)\mathrm e^{-w^2/2}r^{N-1}\mathrm e^{-r^2/2}\mathrm dr\mathrm dw.
$$
The change of variables $(w,r)\to(s,r)$ with $s=w/\sqrt{r^2+w^2}$ yields $w=rs/\sqrt{1-s^2}$ and $\mathrm dw\mathrm dr=(1-s^2)^{-3/2}\mathrm ds\mathrm dr$ hence
$$
\mathrm E(u(\xi))\propto\iint u\left(s\right)\mathrm e^{-r^2s^2/(2(1-s^2))}r^{N-1}\mathrm e^{-r^2/2}(1-s^2)^{-3/2}\mathrm dr\mathrm ds,
$$
which shows that the density of $\xi$ is $f_N$ with
$$
f_N(s)\propto\int\mathrm e^{-r^2s^2/(2(1-s^2))}r^{N-1}\mathrm e^{-r^2/2}(1-s^2)^{-3/2}\mathrm dr=(1-s^2)^{-3/2}\int\mathrm e^{-r^2/(2(1-s^2))}r^{N-1}\mathrm dr.
$$
Using the change of variable $r=(1-s^2)^{1/2}x$ in the last integral shows that it is proportional to $(1-s^2)^{N/2}$ hence $f_N(s)\propto(1-s^2)^{(N-3)/2}$.
I try to reformulate the commentaries of fgp as an answer.
First consider a probability space $(\Omega, \mathcal{A}, \mathbb{P}')$ and a random variable uniformly distributed on $[0,2\pi]$:
$$X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow ([0,2\pi],\mathcal{B}([0,2\pi])).$$
Furthermore consider the parametrization of the unit circle
$$p: [0,2\pi] \rightarrow S^1; \quad x \mapsto e^{i\cdot x}$$
which is continuous.
Now consider the space $(S^1,\mathcal{B}(S^1))$ where $\mathcal{B}(S^1)$ is the $\sigma$-algebra generated by the open sets of $S^1$.
We define $\mathbb{P}$ to be the probability measure induced by the map
$$p\circ X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow (S^1,\mathcal{B}(S^1)); \quad \omega \mapsto e^{i X(\omega)}.$$
Then we have
\begin{equation}
P(Z\in A) = \frac{1}{2\pi i}\int_{A}{\frac{1}{z}}dz \qquad \forall A \in \mathcal{B}(S^1)
\end{equation}
or more generally
$$E[f(Z)]=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz$$
for any measurable, bounded function $f: S^1 \rightarrow \mathbb{R}$
So we must be careful what the measurable space really is. In the case of the "density" in the question we are considering the probability space $(S^1,\mathcal{B}(S^1), \mathbb{P})$ and on this we get can give the value of $\mathbb{P}$ via the above formula.
Hence the above is not a density with respect to the Lebesgue-measure on $\mathbb{C}$, but a way of formulating the value with the help of the parametrization of (or - to be more precise - a contour integral along) the unit circle $S^1$.
Best Answer
It is because the circle is more steep on its left and right edges so a small interval there will contain a lot of arc length. But as you go toward the center of the circle, a small interval will be smaller. Consider an interval of [-1,-.99] vs an interval of [0,.01]. The slope is much higher in the first interval than the second, which is mostly flat, so the first interval will contain more “points,” and hence probability.
Also, Compare the interval [-1,-1/2] vs [-1/2, 0], dividing the left half in two equal interval widths. The first interval contains more arc length than the second. To convince yourself, draw it out. It looks like the first contains about twice as much as the second.
The more arc length, the more “points” there are to choose from.
Two intervals of the same width won’t contain the same arc length unless they are mirrors across the y axis.