If we have, expression (1) with the $\star$ sign used for Hodge star $$\star(d(\alpha))$$ where $\alpha$ is a complex function. We are speaking in 3 dimensions (x,y,z) that is expression(1) can be written as $$\star (\partial_x(\alpha)dx + \partial_y(\alpha)dy+\partial_z(\alpha)dz)$$ this can be written more explicitly as $$\partial_x(\alpha)(dy\wedge dz) + \partial_y(\alpha)(dz\wedge dx) + \partial_z(\alpha) (dx\wedge dy)$$
Okay, my first intention was to take the exterior derivative of expression (1) that is $$d(\star(d(\alpha))$$, now after I expanded it, it seems to me that this is equivalent to: $$d[\partial_x(\alpha)(dy\wedge dz) + \partial_y(\alpha)(dz\wedge dx) + \partial_z(\alpha) (dx\wedge dy)]$$ which pretty much looks like zero to me. Am I thinking in the correct way here?
What are you getting is the Laplacian of $\alpha$ multiplied by the volume element $\text dx\wedge\text dy\wedge\text dz$, i.e. a 3-form, which is dual to the 0-form $\nabla^2\alpha$. Observe that $\delta\text d+\text d\delta = \nabla^2$, where $\delta$ is the codifferential $\star\text d\star$. On 0-forms $\delta$ vanishes, hence $\nabla^2\alpha = \delta\text d\alpha$ in this case.