Möbius values of CNF and DNF lattices of a monotone Boolean function
OK so, more than one year later, here is the answer to this. We'll see Boolean valuations $\nu$ as the set of variables that are mapped to $1$.

We can show that $\mu_\text{cnf}(\hat{0},\hat{1}) = (-1)^k \mu_\text{dnf}(\hat{0},\hat{1}) =  \sum_{\nu \models \phi} (-1)^{|\nu|}$.
In the literature, the quantity $\sum_{\nu \models \phi} (-1)^{|\nu|}$ is also called the Euler characteristic of $\phi$.

I'll only show that $\mu_\text{cnf}(\hat{0},\hat{1}) = \sum_{\nu \models \phi} (-1)^{|\nu|}$, because the proof is similar for $\mu_\text{dnf}(\hat{0},\hat{1}) = (-1)^k \sum_{\nu \models \phi} (-1)^{|\nu|}$.

Let $w \in \mathbb{R}$ be a weight. Define the mass of a valuation $\nu$ under $w$ as $m_w(\nu) = w^{|\nu|} (1-w)^{|\langle k \rangle \setminus \nu|}$, and the mass of $\phi$ under $w$ as $M_w(\phi) = \sum_{\nu \models \phi} m_w(\nu)$. (when $w \in [0,1]$ this can be seen as a probability.)

Define the functions $f,g: L_{\text{cnf}} \to \mathbb{N}$ by:


$f(d_\mathbf{s}) = M_w \left( (\lnot \bigvee_{i \in \mathbf{s}} C_i) \land \bigwedge_{i \in \langle k \rangle \setminus \mathbf{s}} C_i \right)$; in other words the total mass of the valuations that do not satisfy exactly the (disjunctive) clauses $C_i$ for $i \in \mathbf{s}$.
$g(d_\mathbf{s}) = M_w \left( \lnot \bigvee_{i \in \mathbf{s}} C_i \right)$; in other words the total mass of the valuations that do not satisfy at least the clauses $C_i$ for $i \in \mathbf{s}$.


We have $g(x) = \sum_{u \leq x} f(u)$ for all $x \in L_{\text{cnf}}$. So by Möbius inversion formula $f(x) = \sum_{u \leq x} \mu_\text{cnf}(u,x) g(u)$.
    Moreover, for $u = d_\mathbf{s} \in L_\text{cnf}^\phi$, we have that $g(u) = (1 - w)^{|d_\mathbf{s}|}$.
    Now, since $M_w(\phi) = f(\hat{1})$ are the same polynomials in $w$, by identifying the coefficients of $w^{k+1}$ we get that $\mu_\text{cnf}(\hat{0},\hat{1}) = \sum_{\nu \models \phi} (-1)^{|\nu|}$.