Let $\omega$ be a differential $k$-form on $\mathbb{R}^n$. Let $v_1,v_2,...,v_{k+1}$ be $k$ vectors in the tangent space to $\mathbb{R}^n$ at the point $p \in \mathbb{R}^n$.
Given scaling factors $h_1,h_2,...,h_{k+1}$ of the vectors $v_1,v_2,...,v_{k+1}$, we get a parallelepiped $V_h$ defined by the vectors $h_1v_1,h_2v_2,...,h_{k+1}v_{k+1}$. Let $bV_h$ be the oriented boundary of this parallelepiped.
Then the exterior derivative of $\omega$ can be defined by the formula
$$d \omega (p)(v_1,v_2,...,v_{k+1}) = \lim_{h_i \to 0} \frac{1}{h_1h_2...h_{k+1}}\int_{bV_{h}} \omega $$
Notice that this reduces to the definition of the derivative in the case $k=1$ (In this case integration of a zero form over an oriented collection of points is just signed summation).
So I propose the definition:
A $k$-form $\omega$ on $\mathbb{R}^n$ is said to be differentiable if there is a $k+1$-form $d\omega$ such that
$$\int_{bV_{h}} \omega = h_1h_2...h_{k+1} d\omega(p)(v_1,v_2,...,v_{k+1}) + \epsilon|h_1h_2...h_{k+1}|$$
where $\epsilon \to 0$ as the $h_i \to 0$