Ok, let me try to say something intelligent about the 1-dimensional case.
As has already been noted, the normal Noetherian 1-dimensional rings are finite products of Dedekind domains. So what about the non-normal case (say for reduced rings)?
Pass to the normalization
Suppose that $R$ is an excellent purely 1-dimensional reduced ring. Let $R^{\mathrm{N}}$ be the normalization. So we know that $R^{\mathrm{N}}$ is a finite product of Dedekind domains. How do we reconstruct $R$ from it?
It follows that $R$ is always the pullback of a diagram $\{ R^{\mathrm{N}} \to R^{\mathrm{N}}/I \leftarrow S \}$ where $R^{\mathrm{N}}/I$ and $S$ are zero-dimensional (which say we've already classified) and the map $S \to R^{\mathrm{N}}/I$ is injective. To see this, see my answer here and my answer here.
On the other hand, if $A$ is Noetherian normal and purely 1-dimensional, $A/I$ and $B$ are zero dimensional, and $B \to A/I$ is injective, then $\{ A \to A/I \leftarrow B \}$ gives you a 1-dimensional reduced Noetherian ring. So this is at some level a classification (in the second answer, a canonical choice of the $A/I$ and $B$ is given).
On the other hand, if the map $B \to A/I$ is not injective, then this should actually classify (pure) 1-dimensional rings that are generically reduced.
Two dimension and higher
I think is essentially hopeless. :-(