I believe the number is 12.

I will assume the characteristic of the base field is not 2 or 3 so that I can use $\overline{\mathcal{M}}_{1,1} \simeq \mathbf{P}(4,6)$.  Recall that $\mathbf{P}(4,6)$ is constructed by dividing $V = \mathbf{A}^2 \smallsetminus \{ (0,0) \}$ by the weight $(4,6)$-action of $\mathbf{G}_m$.

The line bundle $\mathcal{O}(1)$ (which coincides with the Hodge bundle and generates the Picard group) can be constructed as the equivariant line bundle $V \times \mathbf{A}^1$ on $V$ with $\mathbf{G}_m$ acting by weights $(4,6,1)$.  Then $\mathcal{O}(4)$ has a canonical section $g_4$ and $\mathcal{O}(6)$ has the section $g_6$.  In $\mathcal{O}(12)$ we have the two sections

$1728 g_4^3$

$\Delta = g_4^3 - 27 g_6^2$.

This linear series defines $j : \mathbf{P}(4,6) \rightarrow \mathbf{P}^1$.

Note that $j$ has degree $1/2$ because of the generic automorphism on $\mathbf{P}(4,6)$.  That is, $j_\ast j^\ast$ is multiplication by $1/2$.  Thus the Hodge class $\lambda$ satisfies

$\int_{\overline{\mathcal{M}}_{1,1}} \lambda = \int_{\mathbf{P}(4,6)} c_1(\mathcal{O}(1))
= \frac{1}{12} \int_{\mathbf{P}(4,6)} c_1(j^\ast \mathcal{O}(1)) = \frac{1}{12} \int_{\mathbf{P}^1} j_\ast j^\ast c_1(\mathcal{O}(1)) = \frac{1}{24}$.

Another thing that may be confusing here is that $j$ is generically unramified, so that a local equation for a point in $\mathbf{P}^1$ pulls back under $j$ to a local equation in $\overline{\mathcal{M}}_{1,1}$.  Thus $j^\ast \Delta = \delta$ (if $\Delta$ denotes the boundary in the coarse moduli space and $\delta$ the boundary in the stack).