Let $K/\mathbb{Q}_p$ be a finite field extension and $E/\mathcal{O}_K$ be the local N\'eron model of a CM elliptic curve with CM by $\mathcal{O}_F$ and let $G\subseteq E[p^n]$ be a subgroup over $\mathcal{O}_K$ of order $p^n$ which we assume to be cyclic. We assume that $p$ is inert in $\mathcal{O}_F$ so that $E[p^\infty]$ is a $1$-dimensional $p$-divisible group of height $2$. Hence we can write $$G = \text{Spec}(\mathcal{O}_K[x]/g(x)).$$ Let $s:\mathcal{O}_K\to E$ be the identity section. One can compute from the explicit form of $G$ above that $$s^*\Omega^1_{G/\mathcal{O}_K} = \frac{\mathcal{O}_K}{\prod_{\lambda\in G\setminus \{0\}}\lambda}$$ thus $$\#s^*\Omega_{G/\mathcal{O}_K} = p^{f\sum_{\lambda\in G\setminus \{0\}}v_{\pi_K}(\lambda)}$$ where $f$ denotes the inertia degree of $p$ in $\mathcal{O}_K$.
Since $G\subseteq E[p^n]$, it follows that we can compute $v_{\pi_K}(\lambda)$ from the Lubin-Tate group law on $E[p^\infty]$ since $\lambda$ must also be a zero of that.
We use a Newton polygon for the group law $$[p](X) = pX + X^q$$ where $q = \mathrm{Nm}(\pi_F)=p^2$. Then after a bit of induction, the slopes give the valuation as: $$v_{\pi_K}(\lambda) = \frac{v_{\pi_K}(p)}{q^r - q^{r-1}}$$ for a $\lambda$ of exact order $p^r$.
We let $\Phi(r)$ denote the number of elements of $G$ of exact $p$-order $r$. Then $$\sum_{\lambda\in G\setminus\{0\}}v_{\pi_K}(\lambda) = v_{\pi_K}(p)\sum_{r=1}^n\frac{\Phi(r)}{q^r-q^{r-1}} = v_{\pi_K}(p)\sum_{r=1}^n\frac{p^r-p^{r-1}}{p^{2r}-p^{2r-2}} = \frac{v_{\pi_K}(p)}{p+1}\sum_{r=1}^n\frac{1}{p^{r-1}}.$$
I want a geometric interpretation of this. In particular, if I choose a filtration $$0\subseteq G_1\subseteq\cdots\subseteq G_n = G$$ where each subquotient $G^{(r)}$ is a group of order $p$, then we should have $$\#\log s^*\Omega^1_{G^{(r)}/\mathcal{O}_K} = f\left(\frac{v_{\pi_K}(p)}{p+1}\right)\left(\frac{1}{p^{r-1}}\right)\log p.$$
Is there a way to obtain this without using the formal group law of $E[p^\infty]$? In particular, I am interested in understanding how and why the factor $p^{-(r-1)}$ arises, and whether there is a geometric way of seeing this.
