The bound can be slightly improved to $10^{n/12} \approxeq 1.21153^n$ by the following ten-element covering set for $n = 12$: ${00000000, 11111111, 11000011, 00110011, 11001100, 00111100, 00001111, 11110000, 01101001, 10010110}$ We seem to be observing diminishing returns.

James is correct; I missed the word 'homogeneous' when quickly reading an e-mail containing necessary and sufficient conditions for the Hasse principle to give an algorithmic criterion for whether integer solutions exist.

Indeed, I haven't been able to establish a non-constant lower bound. Any covering set must necessarily contain the words $00\dots 0$ and $11 \dots 1$, and except in the case of $n=3$ these are insufficient. I suspect that we can asymptotically beat $(1 + \epsilon)^n$ for any $\epsilon > 0$.

By the structure theorem for finitely generated modules over a principal ideal domain, every finite Abelian group can be expressed as a direct product of cyclic groups. Do you have a method of implementing cyclic groups as graphs, together with some way of combining graphs so as to multiply their sandpile groups? If so, that solves your question.

"where (as in Fedja's answer) c(n) is the colour of n in the worst-case colouring for van der Waerden's theorem"

Oops, this doesn't work as stated, since $c(1 + \sum_p m_p^2)$ is non-linear and therefore does not yield an arithmetic progression. And any linear function would fail by the same argument as my previous comment.

I think that the sum of squares $c(1 + \sum_p m_p^2)$ should work by Bezout's theorem in algebraic geometry (specifically, a line can intersect a sphere in at most two points). That's not quite as tight as fedja's bound.

@fedja: In your colouring, geometric progressions such as $\{ 3^n, 2 \times 3^{n-1}, 2^2 \times 3^{n-2}, \dots, 2^{n-1} \times 3, 2^n \}$ are monochromatic. Your answer works if we insist that the common ratio is an integer, though.

I'm not sure how the spider relation was discovered, but it certainly stems from a conjecture by John Conway (circa 1980, proved by Ivanov and Norton in 1990) that Y555 quotiented by the spider gives the Bimonster group. This prompted investigation of other Ypqr groups, summarised in a later paper by Ivanov: sciencedirect.com/science/article/pii/S0021869399978821 If the spider relation seems too arbitrary, there is an alternative formulation where the diagram is the 26-vertex incidence graph of the projective plane over F3, and the relations correspond to 'deflating' free 12-gons.

Yes, Coxeter groups are presented by Coxeter-Dynkin diagrams, where each vertex corresponds to a reflection and each pair of points corresponds to a relation $(xy)^n = 1$. When $n = 2$ we omit the edge, when $n = 3$ we draw an unlabelled edge, and otherwise we label the edge with the value of $n$. Some Coxeter groups are infinitely-generated, thus don't correspond to finite diagrams.