This is the first time I try to apply the calculus of differential forms to make some computation so sorry if I say something very silly. My try was the following: $M=\mathbb{R}^3$, and $E\in\mathfrak{X}(M)$ the electric field due to a point charge on the origin we have using the spherical coordinate system
$$E = \dfrac{q}{4\pi\epsilon_0} \dfrac{1}{r^3} \dfrac{\partial}{\partial r},$$
we want to compute the divergence of $E$ using differential forms. Considering then the metric tensor $g = dr^2+r^2d\theta^2+ r^2\sin^2\phi d\phi^2$ we can associate a $1$-form to $E$. Before doing that we are going to use basis $dr$, $r d\theta$, $r\sin\theta d\phi$ so that the metric tensor has the identity matrix at each point. Now, the form associated with $E$, has components $E_i$ given by
$$E_i = g_{ij}E^j,$$
then since $g_{ij} = 0$ for $i\neq j$, $g_{11}=1$ and since $E^j = 0$ for $j \neq 1$ there remains just $E_1 = E^1$ so that we consider the $1$-form
$$E^\flat = \dfrac{q}{4\pi\epsilon_0} \dfrac{1}{r^3}dr,$$
but this is a $1$-form, since we want divergence we need a $2$-form, so we want to take the hodge star operator. In the basis we are working with it maps $dr\mapsto r^2\sin\theta d\theta\wedge d\phi$ so that we have
$$\ast E^\flat = \dfrac{q}{4\pi \epsilon_0} \dfrac{1}{r^3}r^2\sin\theta d\theta\wedge d\phi$$
now we take the exterior derivative of this which by definition is
$$d(\ast E^\flat) = \dfrac{q}{4\pi \epsilon_0}d\left(\dfrac{1}{r^3}\right)\wedge r^2\sin\theta d\theta\wedge d\phi = \dfrac{q}{4\pi \epsilon_0} \dfrac{-3}{r^2}dr\wedge rd\theta\wedge r\sin\theta d\phi,$$
but this isn't right. We know that the result should be zero. So certainly I made some huge mistake there. What's the problem with this computation?
I see two serious errors off the bat. First, in your formula for the metric, you have $\sin\phi$ where you should have $\sin\theta$. But that was a typo you corrected in the next line. More substantively, your formula for $E$ should have $1/r^2$. Then everything works as it should.
EDIT: Oops, one more. You need $d$ of everything once you have the $2$-form!