Let $X \in \mathbb{R}^{n \times p}$ consist of iid $\mathcal{N}(0,1)$. Assume that $n/p$ converges to a positive constant. Denote by $\sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_{\min(n,p)} \ge 0$ the singular values. What's the joint distribution of $$\sigma_1  \sigma_2, \sigma_2  \sigma_3, \sigma_3  \sigma_4, \ldots?$$

$\begingroup$ To find out the joint distribution, I have a post here: mathoverflow.net/questions/158582/… $\endgroup$ – Henry.L Apr 17 '17 at 0:41
The joint distribution of the spacings $x_i=\sigma_i\sigma_{i+1}$, $i=1,2,\ldots m1$, $m=\min(n,p)$, follows from the joint distribution of the singular values $\sigma_i$ ($i=1,2,\ldots m$, ordered from large to small):
$$P(\sigma_1,\sigma_2,\ldots\sigma_m)\propto \prod_{i=1}^m e^{\sigma_i/2}\sigma_i^{\alpha/2}\prod_{i<j}^m (\sigma_i\sigma_j)$$
with the definition $\alpha=\max(n,p)\min(n,p)1$. To obtain the distribution of the spacings you substitute
$$\sigma_i=\sigma_m+\sum_{k=i}^{m1}x_i,\;\;i=1,2,\ldots m1$$
The determinant of this transformation is unity, so you immediately arrive at the joint distribution $P(x_1,x_2,x_{m1},\sigma_m)$ of the $m1$ spacings and the smallest eigenvalue $\sigma_m$. You may or may not want to integrate out the remaining $\sigma_m$. (I would think that setting $\sigma_m\mapsto 0$ should be quite accurate an approximation.)