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)?

<h2>Pass to the normalization</h2>

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][1] and [my answer here][2].  

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 1-dimensional rings that are generically reduced.  

<h2>Two dimension and higher</h2>

I think is essentially hopeless.  :-(


  [1]: http://mathoverflow.net/questions/109395/is-there-a-geometric-intuition-underlying-the-notion-of-normal-varieties/109486#109486
  [2]: http://mathoverflow.net/questions/186406/obtaining-non-normal-varieties-by-pushout/186650#186650