$\newcommand\ep\epsilon\newcommand\de\delta\newcommand\1{\mathbf1}\newcommand\R{\mathbb R}\newcommand\U{\mathscr U}$I think that for any probability distribution $\mu$ over $\R$ there exists a subset $A$ of $\R^n$ of nonzero Lebesgue measure such that each one-dimensional marginal of the uniform distribution $\U_A$ over $A$ is exactly $\mu$.
First, we may consider the case when $\mu$ has a smooth enough density $p$ (with respect to the Lebesgue measure over $\R$) vanishing at $\pm\infty$. Let the desired set $A$ be of the form $$A=A^\de:=\{x\in\R^n\colon \|x-(x\cdot\1)\1/n\|\le\de(x\cdot\1)\},\tag1$$ where $\cdot$ is the dot product over $\R^n$, $\1:=(1,\dots,1)\in\R^n$, $\de$ is a positive function on $\R$ (which should be thought of as small enough), and $\|\cdot\|$ is a norm on $\R^n$ invariant with respect to all permutations of the coordinates. We want the function $\de$ to be such that $$|A^\de_u|=\ep p(u)\tag2$$ for some real $\ep>0$ (which should be thought of as small enough) and all $u\in\R$, where $|A^\de_u|$ is the Lebesgue measure of the $u$-"cross-section" $$A^\de_u:=\{(u_2,\dots,u_n)\in\R^{n-1}\colon(u,u_2,\dots,u_n)\in A^\de\}$$ of $A^\de_u$.
If we can find such a function $\de$, then clearly each one-dimensional marginal of the uniform distribution over $A^\de$ is exactly the distribution $\mu$ with density $p$.
In the case $n=2$, we have $$A^\de=\{(u,v)\in\R^2\colon|u-v|\le c\de(u+v)\}\tag3$$ for some real $c>0$ depending on the choice of the norm $\|\cdot\|$ on $\R^2$ invariant with respect to the permutation of the two coordinates. In this case, (2) is equivalent to the functional equation $$g(u)-g^{-1}(u)=\ep p(u)\quad\forall u\in\R$$ for a function $g\colon\R\to\R$ such that $g(u)>u$ for all real $u$. So, the problem here is to solve this functional equation.
Below are images of Mathematica's work for the set $A^\de$ as in (3) with $c=1/2$ and $\de(t)\equiv\dfrac1{1 + (t - 2)^2}$. In this case, there are explicit expressions for $g(u)$ and $g^{-1}(u)$.
