$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\Spf{Spf}$Background:
If $E$ is a complex-oriented spectrum, then $E^*(K(\mathbb{Z}/p^k,1))$ sits inside a long exact sequence $$E^*(K(\mathbb{Z}/p^k,1)) \to E^*(CP^{\infty}) \stackrel{E^*(p^k)}{\to }E^*(CP^{\infty}),$$ where $E^*(p^k)$ is induced by the multiplication by $p^k$ map on $CP^{\infty}$, i.e., is given by multiplication by $[p^k]_{E}(x)$. This means that $E^*(K(\mathbb{Z}/p^k,1)) = E^*[[x]]/[p^k](x)$. This is a classical computation, that is, for instance, in Hopkins-Kuhn-Ravenel. Using Cartier duality, you get the computation of $E_*(K(\mathbb{Z}/p^k,1))$.
Ravenel-Wilson have computed $K(n)_*(K(\mathbb{Z}/p^k,m))$: it's concentrated in even degrees, and hence is determined by $K(n)_0(K(\mathbb{Z}/p^k,m))$. This is a finite-dimensional Hopf algebra, so we can look at its Dieudonné module, $D(m)$. Then $D(m) = m$ th exterior power of $D(1)$. When $m=1$, we're getting back the same easy computation of $E_*(K(\mathbb{Z}/p^k,1))$ - but when $m=n$, we're getting back a one-dimensional thing. It is also known that the action of the Morava stabilizer group on this top exterior power realizes, as one might expect, the determinant representation.
Lurie's theorem states that if $M \to M_p(h)$ is an etale map from a Deligne-Mumford stack over $\mathbb{Z}_p$ to the moduli of (one-dimensional) $p$-divisible groups of height $h$, then there is a sheaf $O^{\top}$ of $E_{\infty}$-rings on $M$ lifting the structure sheaf of $M$. The proof of this theorem reduces to the fact that Morava $E$-theory is an $E_{\infty}$-ring (since etaleness is essentially saying that the deformation theories of $M$ and $M_p(h)$ are the same).
Question:
So, can we replicate the Ravenel-Wilson computation for $O^{\top}(M)^*(K(\mathbb{Z}/p^k,m))$? The way to go about doing this would be constructing a nice cover of $M$ by affine schemes Spec $R$, evaluating $O^{\top}$ on each of these to get an $E_{\infty}$-ring, $O^{\top}(R)$, and computing $O^{\top}(R)^*(K(\mathbb{Z}/p^k,m))$. Presumably we can then glue all of these local results together via some Bousfield-Kan/Cech-spectral sequence.
Alternate motivation
There's another motivation for doing this: the $K(n)$-local Picard group has no "exotic" elements for $p\gg n$. There are already exotic elements at $n=1$ and $p=2$; Heard-Stojanoska have shown that $P := S^{-2} I_1$ is an exotic element of Pic$_1$, where $I_1 = L_{K(n)} S I_Z$. In particular, for $p\gg n$, $S[\det] = S^{n-n^2} L_{K(n)} I_{\mathbb{Q}/\mathbb{Z}}$. When $p\gg n$ isn't satisfied, this is probably not true. For instance, Heard-Stojanoska's spectrum $P$ gives an example. I think that to show $S[\det]$ and $S^{n-n^2} L_{K(n)} I_{\mathbb{Q}/\mathbb{Z}}$ aren't the same at $n=2$ and $p=3$, we need to understand $(L_{K(2)} \mathit{tmf})_*(K(\mathbb{Z},3))$. Doing a computation like the one suggested above will answer this question.
Approach 1:
One approach to that specific computation that doesn't involve a Bousfield-Kan spectral sequence: Hill's The $3$-local tmf-homology of $BS_3$ establishes a spectral sequence (see Corollary 1) $\Ext_{A(1)_* (x) E(a_2)}(F_3, H_*(X)) \Rightarrow \mathit{tmf}_*(X)$, where $ A(1)_*$ is dual to the subalgebra of the Steenrod algebra generated by the Bockstein and $P^1$, so that $A(1)_* = F_3[xi_1]/(xi_1^3) (x) E(\tau_0, \tau_1)$. Presumably we can compute $\mathit{tmf}_*(K(Z,3))$ from this spectral sequence. We can impose a grading so that $A(1)_*$ has filtration zero and a_2 has filtration 1. Letting $A = A(1)_* (x) E(a_2)$, there's an associated graded spectral sequence $\Ext_{Gr(A)}(F_3, H_*(X)) \Rightarrow \Ext_A(F_3, H_*(X))$. In turn, a Cartan-Eilenberg spectral sequence gives rise to another spectral sequence converging to the $E_2$-page of the associated graded sseq, namely $\Ext_{E(a_2)}(F_3, \Ext_{A(1)_*}(F_3, H_*(X))) \Rightarrow \Ext_{Gr(A)}(F_3, H_*(X))$. The first step would therefore be to write $H_*(K(Z,3); F_3)$ as a $A(1)_*$-comodule.
Approach 2:
Perhaps that's not the most efficient method, though. Instead, recall that $L_{K(2)}$ TMF is the product of $E_2^{h \Aut(C)}$ over all supersingular curves $C$ over $F_p$-bar. It follows that, since the Tate spectra of finite subgroups of the Morava stabilizer group vanish, there is an equivalence (everything is a completed smash product here) $L_{K(2)} TMF \wedge K(Z, 3) = \prod_{s.s. C/F_p-bar} (E_2 \wedge K(Z, 3))^{h \Aut(C)}$. So, to compute the completed ($K(2)$-local) TMF-homology of $K(Z, 3)$, we need to compute the HFPSS $H^{*}(\Aut(C); (E_2)^{_*} K(Z,3)) \Rightarrow \pi_*((E_2 \wedge K(Z, 3))^{h \Aut(C)})$. We know $(E_2)^{_*} K(Z,3)$, so this is reduced to a (hard) computation.
If we're to try to do the computation via a Cech spectral sequence for a chosen cover of M, we need to understand the Landweber-exact cohomology of $K(\mathbb{Z}/p^k,m)$. One way to go about doing this would be to describe $MU^*(K(\mathbb{Z}/p^k,m))$ as a $MU^*$-module.
Approach 3:
Here's another approach that's probably the one to take: because $M \to M_p(h)$, if $\Spf(R) \to M$ is an affine etale cover, we must have $R$ Noetherian, and there's a maximal ideal $m$ of $R$ such that $R^{_m}$ is isomorphic to $W(k)[[u_1,...,u_{h-1}]]$, where $k = R^{_m}/m$. Can we somehow use this to bootstrap from the Morava $E$-theory computation of $K(\mathbb{Z}/p^k,m)$? I think it's true that the completion of $O^{\top}(R)$ at that maximal ideal is equivalent to Morava $E$-theory at height $h$.
If we want to assemble all these local computations together, we need to know how the maximal ideal varies as we move from $\Spf(R)$ to $\Spf(R')$ over $M$. To see how this maximal ideal comes about, look at Lemma 8.1.6 of Behrens-Lawson's Topological Automorphic Forms.
Classically, if $M$ is a finitely generated $R$-module, and $I$ is a fixed ideal of $R$, then $M^{_I} = M (x)_R R^{_I}$. We can assume that the $E_{\infty}$-rings that we're considering are even periodic, for now; then the same is true in the derived context - see Corollaries 7.3.6.4 and 7.3.6.6 of SAG. As a consequence, we have $O^{\top}(R)^{_m} = E_h$, and then $E_h \wedge K(\mathbb{Z}/p^k,m) = O^{\top}(R)^{_m} \wedge K(\mathbb{Z}/p^k,m)$. But now, $(O^{\top}(R) \wedge K(\mathbb{Z}/p^k,m))^{_m} = (O^{\top}(R) \wedge K(\mathbb{Z}/p^k,m)) \wedge_{O^{\top}(R)} O^{\top}(R)^{_m}= K(\mathbb{Z}/p^k,m) \wedge O^{\top}(R)^{_m}$. So, taking homotopy, we get an isomorphism $(O^{\top}(R)_*K(\mathbb{Z}/p^k,m))^{_m} = (E_h)_*K(\mathbb{Z}/p^k,m)$. Can we identify $O^{\top}(R)_*K(\mathbb{Z}/p^k,m)$ from this, given that we understand its $m$-completion?