Take $X=(0,1)^n.$ Fix points $p,q$ s.t. $\text{dist}_n(p,q)=\sqrt{n}.$
Problem: Classify analytic regular foliations of $X$ with $(n-1)-$dim. leaves which are topologically $(0,\sqrt{n})\times S^{n-2} $ accumulating to $p,q.$
The problem is trivial until dim. $2,$ and I have found one (real analytic foliation) solution in $n=2$ dim. but have not completely classified for $n=2.$
I conjecture that there are no solutions for $n\gt2$ and I would be shocked if there was a solution.
My solution for $n=2$:
Foliation $F_s=\big\{\log x \log y=s: s\in (0,\infty) \big\}$ which satisfies $\cup_s F_s=X,$ is clearly analytic and regular, and satisfies the topological constraint as well as the accumulation criterion. In this case, note that $p=(0,1)$ and $q=(1,0).$
I feel like there should be a more elegant approach here. My solution does not let one see the global classification picture and is only an isolated example that happens to work.
Perhaps the problem is easier in the complex setting.
Edit 3/19/2023:
I will relax the conditions. Let $n=3.$
Take $X=(0,1)^n.$ Fix points $p,q$ s.t. $\text{dist}_n(p,q)=\sqrt{n}.$ Construct a smooth regular foliation of $X$ with $(n-1)-$dim. leaves which are topologically $(0,\sqrt{n})\times S^{n-2} $ accumulating to $p,q.$
I think this is a much more tractable problem. I am only looking for a construction in dimension $3$ now, where I've replaced "analytic" with "smooth."
I am including the following related question I posted $10$ months ago (link below), because it is an attempt at a solution (in the smooth case) for any dimension $n.$ This method leverages distribution theory and imagines the leaves as arising from integrating some continuous bivariate probability distribution. In this way a single leaf is precisely the distribution when all the parameters are fixed. What's interesting about this method is that you get the topological condition for sufficient constraints on the probability distribution.
Since the definition in the linked post below includes $p,q$ in the leafs, then deleting $p,q$ would match up with my definition in this post.