**EDIT:** aside from sign errors, I had realized that the notation I used implied something completely different from what I had derived (I don't think the resulting expressions had a definition!). It should be fixed now, as well as slightly less step-by-step, so the reader can understand the process of the derivation.

Let $S=\displaystyle\int_{\Omega} \mathcal{L}$ be the action for a Lagrangian D-form $\mathcal{L}=\mathcal{L}(\omega,\star\omega,d\omega,d\star\omega,\star d\omega, \star d\star\omega)$ where D is the dimenstion of space(time), $\omega$ is a q-form and $\star$ is the Hodge star.

Defining derivatives with respect to a form as $\partial_{\alpha_i} (\alpha_0 \wedge \ldots \wedge \alpha_n) = \alpha_0 \wedge \ldots \wedge \alpha_i \! \! \! \! \! \Big/ \wedge \ldots \wedge \alpha_n$, and denoting the position of the form within the wedge products as $l(\alpha_i)=i$, the variation of the action becomes

$$\frac{d}{d\epsilon}S[\omega+\epsilon\chi]\Big|_{\epsilon=0}$$

$$=\int_\Omega (-1)^{l(\omega)}\chi \wedge \partial_\omega\mathcal{L} + (-1)^{l(\star\omega)}\star\chi \wedge \partial_{\star\omega}\mathcal{L} + (-1)^{l(d\omega)}d\chi \wedge \partial_{d\omega}\mathcal{L}\\ + (-1)^{l(d\star\omega)}d\star\chi \wedge \partial_{d\star\omega}\mathcal{L} + (-1)^{l(\star d\omega)}\star d\chi \wedge \partial_{\star d\omega}\mathcal{L} + (-1)^{l(\star d\star\omega)}\star d\star \chi \wedge \partial_{\star d\star\omega}\mathcal{L}$$

Exterior derivatives can then be "donated" to the Lagrangian from the product rule $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^q \alpha \wedge d\beta$, and likewise for the Hodge star operator given the identities $\alpha\wedge\star\beta =\langle\alpha,\beta\rangle\nu=\star\alpha\wedge\beta$ (if the order of alpha and beta sum to D) and $\star \star \alpha = (-1)^{q(D-q)}s\alpha$.

$$\int_\Omega \left(\alpha\partial_\omega\mathcal{L} + \beta\star\partial_{\star\omega}\mathcal{L} + \gamma d\partial_{d\omega}\mathcal{L} + \delta\star d\partial_{d\star\omega}\mathcal{L} + \zeta d\star\partial_{\star d\omega}\mathcal{L} + \xi \star d\star\partial_{\star d\star\omega}\mathcal{L}\right)\wedge\chi \\ + d\left[\left(\gamma^* \partial_{d\omega}\mathcal{L} + \zeta^*\star\partial_{\star d\omega}\mathcal{L} \right)\wedge\chi +\left(\delta^*\partial_{d\star\omega}\mathcal{L} + \xi^*\star\partial_{\star d\star\omega}\mathcal{L}\right)\wedge\star\chi\right]$$

To avoid clutter I have written sign prefactors with greek letters. They can be found using the exterior derivative and hodge star's identities.

Since $\chi$ is a symmetry of the action, it follows that $\delta\mathcal{L}[\omega + \epsilon \chi] = \delta\mathcal{L}[\omega] + d\Lambda$ for a (D-1)-form $\Lambda$. On shell, then, requiring $\delta S=0$, the conservation law manifests:

$$dJ=0$$

$$J=\left(\gamma^* \partial_{d\omega}\mathcal{L} + \zeta^*\star\partial_{\star d\omega}\mathcal{L} \right)\wedge\chi +\left(\delta^*\partial_{d\star\omega}\mathcal{L} + \xi^*\star\partial_{\star d\star\omega}\mathcal{L}\right)\wedge\star\chi-\Lambda$$