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 $P_h$ defined by the vectors $h_1v_1,h_2v_2,...,h_{k+1}v_{k+1}$. Let $bP_h$ be the oriented boundary of this parallelepiped.
Then the exterior derivative of $\omega$ can be defined by the formula
$$d \omega (p)(v_1,v_2,...,v_{k+1}) = \lim_{h_i \to 0} \frac{1}{h_1h_2...h_{k+1}}\int_{bP_{h}} \omega $$
Notice that this reduces to the formula for 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:
Let given $k+1$ vectors $v_1,v_2,...,v_{k+1}$, let $P$ be the oriented parallelepiped spanned by these vectors.
A $k$-form $\omega$ on $\mathbb{R}^n$ is said to be differentiable if there is a $k+1$-form $d\omega$ such that
$$\int_{bP} \omega = d\omega(p)(v_1,v_2,...,v_{k+1}) + \epsilon|Vol(P)|$$
where $\epsilon \to 0$ as the $v_i \to 0$
Thinking about this a bit more, what if we try to replace the integration with just evaluation?
Just to see an example play out first, consider a one form $\omega$ on $\mathbb{R}^n$. It seems reasonable to demand the following approximation (to be made precise shortly):
$d\omega(p)(v_1,v_2) \approx \omega(p)(v_1)+\omega(p+v_1)(v_2) - \omega(p+v_2)(v_1) - \omega(p)(v_2)$
This is obtained by just making a picture of the parallelogram spanned by $v_1$ and $v_2$ and taking signed sum of the boundary.
Example:
If we take a one form $\omega = fdx +gdy$ on $\mathbb{R}^2$ then we should have
$$\begin{align}
d\omega(p)(v_1,v_2) &\approx \omega(p)(v_1)+\omega(p+v_1)(v_2) - \omega(p+v_2)(v_1) - \omega(p)(v_2)\\
&=f(p)dx(v_1)+g(p)dy(v_1) + f(p+v_1)dx(v_2)+g(p+v_1)dy(v_2)-f(p+v_2)dx(v_1) - g(p+v_2)dy(v_1) - f(p)dx(v_2)-g(p)dy(v_2)\\
&=[f(p+v_1)-f(p)]dx(v_2)-[f(p+v_2)-f(p)]dx(v_1)+[g(p+v_1)-g(p)]dy(v_2)-[g(p+v_2)-g(p)]dy(v_1)\\
&\approx df(p)(v_1)dx(v_2)-df(p)(v_2)dx(v_1) + dg(p)(v_1)dy(v_2)-dg(p)(v_2)dy(v_1)\\
&=(df \wedge dx)(v_1,v_2)+(dg \wedge dy)(v_1,v_2)\\
&=(df \wedge dx+dg \wedge dy)(v_1,v_2)
\end{align}$$
This is, of course, the usual formula for the exterior derivative.
So I now propose the following definition:
A $k$-form $\omega$ is said to be differentiable if there is a $k+1$ form $d\omega$ such that
$$\sum_{i=1}^{k+1} (-1)^i(\omega(p+v_i)-\omega(p))(\widehat{v_i}) = d\omega(p)(v_1,v_2,...,v_{k+1}) + \epsilon \left| \textrm{Vol}(P) \right|$$,
where $\epsilon \to 0$ as $v_1,v_2,...,v_{k+1} \to 0$, $P$ is the parallelepiped based at $p$ and defined by the vectors $v_1,v_2,...,v_{k+1}$. $\widehat{v_i}$ is the ordered list $v_1,v_2,...,v_{k+1}$ with $v_i$ left out.
The expression on the LHS of the expression is "$\omega$ applied to the oriented boundary of $p$".
In the case that $\omega$ is differentiable according to the definition we have given above, we can obtain a formula for the exterior derivative which is similar in spirit to the formula for the derivative.
Namely, we have
$$
\begin{align}
d\omega(p)(v_1,v_2,...,v_{k+1}) &= \sum_{i=1}^{k+1} (-1)^i(\omega(p+v_i)-\omega(p))(\widehat{v_i}) - \epsilon \left| \textrm{Vol}(P) \right|
\end{align}
$$
so
$$
\begin{align}
d\omega(p)(hv_1,hv_2,...,hv_{k+1}) &= \sum_{i=1}^{k+1} (-1)^i(\omega(p+hv_i)-\omega(p))(h\widehat{v_i}) - \epsilon \left| \textrm{Vol}(P_h) \right|\\
h^{k+1}d\omega &= h^k \sum_{i=1}^{k+1} (-1)^i(\omega(p+hv_i)-\omega(p))(\widehat{v_i})- \epsilon h^{k+1}\left| \textrm{Vol}(P_1) \right|\\
d\omega &= \sum_{i=1}^{k+1} (-1)^i\left[\frac{\omega(p+hv_i)-\omega(p)}{h}\right](\widehat{v_i})- \epsilon \left| \textrm{Vol}(P_1) \right|
\end{align}
$$
Taking limits of both sides as $h \to 0$, we have
$$d\omega = \lim_{h \to 0} \sum_{i=1}^{k+1} (-1)^i\left[\frac{\omega(p+hv_i)-\omega(p)}{h}\right](\widehat{v_i})$$
In the case that $\omega$ is expressed in the form $\displaystyle\sum_{|I| = k} f_I dx_I$ (using multi index notation), you could further reduce this formula to the usual $\displaystyle\sum_{|I| = k} d(f_I) \wedge dx_I$
Best Answer
The funny thing is that I got almost the same question from one of my student last semester. The idea was to neglect both the linear and quadratic parts of the hypothetical expansion (without loss of generality, we may assume they are zero), and focus on $ \epsilon \|h\|^2$. We came to the following (obvious) one-dimensional counterexample:
$$f(h) = h^3 \sin(\frac{1}{h})$$
for $h \not= 0$, and: $$f(0) = 0$$