If you are in characteristic zero and $G$ is unipotent, then the exponential map is a $G$-equivariant isomorphism of algebraic varieties $\mathfrak{g} \longrightarrow G$, and therefore $\mathfrak{g}//G \cong G//G$ in this case.
Also, in characteristic zero but now letting $G$ be any algebraic group, the exponential map is a $G$-equivariant isomorphism from the completion of $\mathfrak{g}$ at $0$ to the completion of $G$ at $e$. I think the completions of $\mathfrak{g}//G$ and $G//G$ at the corresponding maximal ideals should be isomorphic, but Friedrich Knop has left a skeptical comment below, so I'll think about it more.
$\DeclareMathOperator\PSL{PSL}$In comments (1 2) to skd's answer, I pointed out that $\PSL_3//\PSL_3$ is the singular variety $\operatorname{Spec} k[x,y,z]/\langle xy=z^3 \rangle$. Here we have $x=e_1^3$, $y=e_2^3$ and $z=e_1 e_2$ where $e_1$ and $e_2$ are the first and second elementary symmetric functions of the eigenvalues of a matrix in $SL_3$. Note that the identity corresponds to $e_1=e_2=3$, $(x,y,z) = (27,27,9)$, which is a smooth point of $\operatorname{Spec} k[x,y,z]/\langle xy=z^3 \rangle$; the singularity is at $e_1=e_2=0$, $(x,y,z)=0$, which corresponds to a matrix with eigenvalues $(1,\omega, \omega^2)$ for a primitive root of unity $\omega$. So the completion is just a power series ring in two variables.