Let $\Sigma_{n}$ denote the symmetric group on $n$ letters. $\Sigma_{n}$ acts on unordered pairs $\{i,j\}$ via $\sigma(i,j)=\{\sigma(i),\sigma(j)\}$. Let $e_{\{i,j\}}$ be a basis for $\mathbb{R}^{N}$ where $N=\binom{n}{2}$ and $\Sigma_{n}$ acts by permuting coordinates: $\sigma \cdot e_{\{i,j\}}=e_{\sigma^{-1}\{i,j\}}$. Let $\rho_{n}$ denote this representation.
For the usual symmetric product it isn't hard to write down an explicit homeomorphism $\mathbb{R}^{n}/\Sigma_{n} \cong \mathbb{R} \times \mathbb{R}_{\ge 0}^{n-1}$: given $\left[x\right] \in \mathbb{R}^{n}/\Sigma_{n}$, reorder $x$ using the order on $\mathbb{R}$ so that $t_{1} \le \ldots \le t_{n}$ and $t_{i}=x_{\sigma(i)}$ for some $\sigma \in \Sigma_{n}$. Then $\left[x\right] \mapsto (t_{1},t_{2}-t_{1},\ldots,t_{n}-t_{1})$.
I'd like to understand the quotient $\mathbb{R}^{N}/\Sigma_{n}$ as a topological space or an orbifold. One thought was to try and first break up $\rho_{n}$ into irreducibles, then analyze the orbits on various pieces. The representation looks like $\bigwedge^{2}\mathbb{R}^{n}$, however here $(i \, j)\cdot e_{\{i,j\}}=e_{\{i ,j\}}$ whereas $(i \, j)\cdot e_{i} \wedge e_{j} = - e_{i} \wedge e_{j}$.
My guess is that $\mathbb{R}^{N}/\Sigma_{n}$ should look similar to $\mathbb{R}^{N}/\Sigma_{N}$, except now the fundamental domain looks like $\left[ \Sigma_{N}:\Sigma_{n}\right]$ copies of the fundamental domain in $\mathbb{R}^{N}/\Sigma_{N}$ glued together along strata. However, I'd still be looking for an explicit map into Euclidean space.