I understand that, in a uniform lattice of $vcd = n$, the kernel of an epimorphism to $\mathbb{Z}$ has $vcd = n -1$ -- it is strictly smaller than $n$ by Poincaré duality, and at least $n-1$ by looking at the spectral sequence. Could you elaborate on why this suffices to obtain all intermediate dimensions? Can one iterate this process somehow?

@Z.M yes, but this was just for the sake of the argument - when the coefficients and solutions are both polynomials we can just express everything in terms of the Weyl algebra. For analytic solutions one just changes all the above in accordance with the proper ring of differential operators $\mathcal{D}$ and module of coefficients. The point is that even in the polynomial case the answer is not clear to me.

In fact, if $e(M)$ denotes the degree of the Hilbert polynomial of the solution space and $d(M)$ the dimension of the characteristic variety, the statement I'm looking after would be something like this: $e(M) + n = d(M)$. For holonomic $M$ we have $d(M) = n$ and the solution space is finite dimensional, so $e(M) = 0$. For $M$ presented by killing only one derivative $\partial_i$, the characteristic variety is a hyperplane, so $d(M) = 2n -1$, and the solution space is a space of polynomials in $n-1$-variables, so $e(M) = n-1$.

I've been thinking a bit more about this, and if $R$ denotes the coordinate ring $\mathbb{R}[x,y,\partial_x,\partial_y]/J$, then the graded object associated to the degree function on the solution space embeds in the group $\operatorname{Hom}_{\operatorname{gr} \mathcal{D}}(R, \mathbb{R}[x,y])$ of graded homomorphisms. But playing around with some examples, I don't think this says much, as for $T = \partial_i$ this graded $\operatorname{Hom}$ group is just isomorphic to $\mathbb{R}[x,y]$!

@Z.M restricting from power series to polynomials, to say that $\mathcal{D}/\mathcal{D}T$ is regular holonomic, it means that the graded ideal $J = \operatorname{gr} \mathcal{D}T$ of the polynomial ring $\mathbb{R}[x,y,\partial_x,\partial_y]$ is radical and the associated characteristic variety is 2-dimensional in $\mathbb{R}^4$ (ignoring any issue coming from $\mathbb{R}$ not being algebraically closed). How does this relate to the solution space $\operatorname{Hom}(\mathcal{D}/\mathcal{D}T, \mathbb{R}[x,y])$? Can one read the dimension from this vector space?

@Kimball I see, so what I should search for is something along the lines of "classification of irreducible automorphic representations" of $\operatorname{PGL}_2(\mathbb{R})$ and $\operatorname{PGL}_2(\mathbb{C})$?

Thanks for the answer! The argument seems to reflect geometrically what is going on with Cassels' proof, which uses a mix of Noether's normalization and Hensel's lemma. The primes avoided by the localization process are detected in the Noether normalization, the existence of an $\mathbb{F}_p$ point is again given by analytic methods and the fact that to embed the completion of $R$ in $\mathbb{Z}_p$ it suffices to find small algebraically independent units is shown using Hensel's lemma.

Two points: (1) it is clear that $R$ must be a commutative domain, I forgot to add that hypothesis. Otherwise, Noether's normalization either doesn't apply or only gives us integrality. (eg. $R = \mathbb{Z}[x]/(x^2)$). (2) I was unware of Cassels' embedding theorem, which proves a way stronger statement: if $R$ is a finitely generated commutative domain, for infinitely many primes $p$ there is an embedding of $R$ into $\mathbb{Z}_p$ itself. I am unsure if I should keep the question or edit it.