Let's see what happens when we interpret these sums pointwise. For example, $\partial I^2$ is an 1-chain given by
$$\partial I^2 = -I^2_{(1,0)} + I^2_{(1,1)} + I^2_{(2,0)} - I^2_{2,1}$$
Now evaluating this "sum" at a point $x \in [0,1]$ gives
$$\partial I^2 = -(0,x) + (1, x) + (x, 0) - (x, 1) = (1, -1) $$
Try the same calculation for $\partial I^3$ and you'll get $(1,-1,1)$. Clearly this doesn't make sense.
To see how Spivak "summed" the $(n-2)$-chain in his proof of $\partial^2=0$, let's take $\partial^2I^2$ as an example. Now we must sum the boundaries of each $I^2_{(i, \alpha)}$ with the correct orientation.
So, for instance,
\begin{align}
\partial (-I^2_{(1,0)})(0)
&= -[-(I^2_{(1,0)})_{(1,0)}(0) + (I^2_{(1,0)})_{(1,1)}(0)] \\
&= -[-I^2_{(1,0)}(I^1_{(1,0)}(0)) + I^2_{(1,0)}(I^1_{(1,1)}(0))] \\
&= -[-I^2_{(1,0)}(0) + I^2_{(1,0)}(1)] \\
&= -[-(0,0) + (0,1)]\\
&= (0,0) + - (0,1)
\end{align}
We don't add these: they are just "oriented" points in $\mathbb{R}^2$.
Similarly, we get
\begin{align}
\partial(I^2_{(1,1)})(0) &= -(1,0) + (1,1)\\
\partial(I^2_{(2,0)})(0) &= -(0,0) + (1,0)\\
\partial(-I^2_{(2,1)})(0) &= (0,1) + -(1,1)
\end{align}
Each of the four corners of the square are appear twice with opposite signs.
For instance,
$(0,1) = (I^2_{(1,0)})_{(1,1)} = (I^2_{(2,1)})_{(1,0)}$ (this is the statement $(I^n_{(i,\alpha)})_{(j,\beta)} = (I^n_{(j+1,\beta)})_{(i,\alpha)}$ in Spivak's proof). But it's orientation is negative in the former and positive in the latter. So they cancel each other out - they are not "added".
To summarize, we don't add two different n-cubes in an n-chain. We add the coefficients of the same n-cube if it appears more than once.
No, the correct generalization involves a "higher-dimensional analogue" of simple connectivity. A closed $2$-form will be exact if the region has no "two-dimensional holes." The correct notion is homology or cohomology. If the $k$th cohomology $H^k(X,\Bbb R)$ (which can be computed topologically or in terms of differential forms) vanishes, then every closed $k$-form is exact. Indeed, if the $1$st cohomology vanishes, then every closed $1$-form is exact; this condition is in fact weaker than simple connectivity. (Rather than needing two paths to be homotopic, it is enough for them to be homologous.)
To answer your question specifically, $\Bbb R^3-\{0\}$ is simply connected, but the $2$-form
$$\omega = \frac{x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy}{(x^2+y^2+z^2)^{3/2}}$$
is closed but not exact. (Its integral over any sphere centered at the origin, for example, is $4\pi$.)
Best Answer
For the 2-sphere, it isn't hard to show it directly. However, in general, you'll need more powerful tools. A manifold $M$ is simply-connected if it is path-connected and if $\pi_1(M) = 1$. The standard tool to compute $\pi_1$ is the van Kampen theorem.