How rough can differential form, manifold and chain be for Stokes theorem to hold?

Stokes' theorem holds for manifolds with corners. The following is Theorem $16.25$ on page $419$ in John M. Lee's book $\textit{Introduction to Smooth Manifolds}$.

$\textbf{Theorem}$ (Theorem $16.25$, [Lee]). Let $M$ be an oriented smooth $n$-manifold with corners, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then $$ \int_M d\omega = \int_{\partial M}\omega. $$

The author proves the above theorem in the book. Furthermore, Lee also discusses and proves Stokes' theorem for surface integrals (see Theorem $16.34$ on page $427$).

$\textbf{Theorem}$ (Theorem $16.34$, [Lee]). Let $M$ be an oriented Riemannian $3$-manifold with or without boundary, and let $S$ be a compact oriented $2$-dimensional smooth submanifold with boundary in $M$. For any smooth vector field $X$ on $M$, $$ \int_S \langle \text{curl } X,N \rangle_g \:dA = \int_{\partial S}\langle X,T\rangle_g \: ds, $$ where $N$ is the smooth unit normal vector field along $S$ that determines its orientation, $ds$ is the Riemannian volume form for $\partial S$ (with respect to the metric and orientation induced from $S$), and $T$ is the unique positively oriented unit tangent vector field on $\partial S$.

In Lang's Real Analysis you'll find Stokes's Theorem stated for a $C^2$ manifold and $C^1$ form $\omega$. He goes on to discuss what to do with singularities.

Although Lang asserts existence of a $C^2$ partition of unity using the $C^2$ hypothesis, a $C^1$ partition of unity suffices. However, as @MatheinBoulomenos pointed out to me in chat, you need a $C^2$ chart in order to be sure that the pullback of the $C^1$ form (bumped off by the $C^1$ partition of unity) is still a $C^1$ form. (Lang is very sloppy about paying attention to details in this regard.) Then it's the usual computation.