Higher (chain) homotopies I am aware of this question, which unfortunately doesn't help me enough.
Recall that a (chain) homotopy between maps $f, g\colon X_\bullet\to Y_\bullet$ of chain complexes is a collection of maps $h_\bullet\colon X_\bullet \to Y_{\bullet+1}$ such that $\partial h + h\partial = f-g$. It is reasonable to call this a chain homotopy, as homotopies of continuous maps induce homotopies of singular chain complexes.
Now, how does this generalize to higher dimensions?
Top: I guess that we are interested in homotopies of homotopies, i. e., for two homotopies $H, K\colon f\rightsquigarrow g$ which are maps $I\times X\to Y$, we want to find a map $\Upsilon\colon I^2\times X$ such that $\Upsilon_{\{0\}\times I\times X}=H$ and $\Upsilon_{\{1\}\times I\times X}=K$. Please correct me if I am wrong.
Ch: I have no clue what the right notion for chain complexes is supposed to be. At least, $h$ from above wasn't supposed to be a chain map; I guess it's incorrect to ask for an $\upsilon$ such that $h - k = \partial \upsilon + \upsilon \partial$.
One guess: I am trying to study this introduction of $A_\infty$-algebras. Spelling out the relations for $n=4$ makes me wonder if some formula involving
$\upsilon\partial - \partial\upsilon - \upsilon$$
does the job. 
 A: One way to view this ( I don't know if it's the standard notion of higher homotopies in chain complexes) is as follows :
First recall that there's a chain complex $Hom(X,Y)$, defined by $Hom(X,Y)_n = \prod_{m\in\mathbb Z}Hom(X_m, Y_{n+m})$ and the differential is $\partial f = \partial_Y \circ f - (-1)^{|f|}f\circ \partial_X$.
It's easy to  check that chain maps are precisely $0$-cycles in $Hom(X,Y)$, and that a homotopy from $f$ to $g$ is just an element $h$ in degree $1$ with $\partial h = f-g$. 
Now assume $k$ is another such homotopy; then $\partial(h-k) = 0$, and so we can be tempted to define a higher homotopy to be simply an element $H$ in degree $2$ with $\partial H = h-k$, etc. for higher ones.
Now there's another point of view on the same idea. Suppose for instance that $f=g=0$ (so we're looking at self-homotopies of the $0$ map), then $\partial h =0$ as well, this means $\partial_Y\circ h + h\circ \partial_X = 0$.
Now if you shift $X$ by one degree, so take $(\Sigma X)_n = X_{n-1}$ and take $\partial_{\Sigma X} = -\partial_X$, you get that $h: \Sigma X\to Y$ is actually a chain map ! 
In fact more generally, $Hom(\Sigma X,Y)_n = Hom(X,Y)_{n+1}$ and you can check that this identification actually respects $\partial$, so an $H$ as before just becomes an honest homotopy between $h$ and $k$.
But this is exactly what we would want ! Indeed, in topological spaces, $Map_*(\Sigma X,Y)$ is just the space of (pointed) homotopies between the null map and itself, so homotopies in it are just higher homotopies between homotopies between the null map and itself. 
So our $H$ plays the same role (if you accept that the two $\Sigma$'s introduced play the same role, but there are good reasons to believe that) as higher homotopies topologically.
But once you spot that, the generall pattern is not hard to notice : a homotopy between $h,k$ will always be an $H$ with $\partial H = h-k$, no matter the degree. 
Now these are "pointed" homotopies in a sense, and there's another story you could tell with unpointed homotopies. Indeed, a homotopy can be defined in terms of a certain interval object, just as in topology. 
Take $I= \dots \to 0\to  \mathbb Z\to \mathbb{Z\oplus Z}\to 0\to \dots$ where $\mathbb{Z\oplus Z}$ is in degree $0$, and the nontrivial differential is $1\mapsto (-1,1)$. This is an algebraic model for the interval (the interval has two points, so two zero simplices, and one edge between them - that edge ges from $0$ to $1$, and that explains the minus sign in there). 
There are two maps from $\mathbb Z[0]$ to $I$, one denoted $i_0$, the inclusion into the left factor, and $i_1$, the inclusion into the right factor. 
It's a good exercise to check that the data of a homotopy between $f,g : X\to Y$ is the same as a map $X\otimes I\to Y$ such that the composite $X\cong X\otimes \mathbb Z[0] \overset{1\otimes i_0} \to X\otimes I\to Y$ (resp. $1\otimes i_1$) is $f$ (resp. $g$); up to a few sign considerations that are easy to fix. 
Using this, it's easy to define a notion of higher homotopies : just do like for spaces : a homotopy between homotopies will just be a map $X\otimes I \otimes I\to Y$ satisfying certain things. If you want the "endpoints" to be fixed, you'll have to add that in your definition, but hopefully the idea should be clear at that point. 
It can be an interesting exercise to see how the two points of view relate (I couldn't tell you a precise statement on the top of my head, so I'll leave that to you as well !)
