The exterior algebra formed from a vector space $V$ with vector basis $\{e_1,e_2,e_3\}$ will have basis
$$\begin{align} &\Delta^0 V=\langle 1 \rangle\\ &\Delta^1 V=\langle e_1,e_2,e_3 \rangle\\ &\Delta^2 V=\langle e_1\wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \rangle\\ &\Delta^3 V=\langle e_1\wedge e_2 \wedge e_3 \rangle\\ \end{align}$$
I presume...
The alternating algebra, also called the exterior algebra, $\Delta^* V$ is a $2^n$ dimensional algebra. In Wolfram Language, an element of the alternating algebra can be represented by an $n$-nested binary list. For example,
$$\{\{\{\color{blue}1,\color{red}2\},\{0,0\} \},\{\{\color{orange}3,0\},\{\color{magenta}4,\color{brown}5\}\}\}$$
represents
$$\begin{align} \color{blue} 1 \;e_1 \wedge e_2 \wedge e_3\quad+\\ \color{red}2\; e_1 \wedge e_3\quad+\\ \color{orange}3\;e_2\wedge e_3\quad +\\ \color{magenta}4\; e_3\quad +\\ \color{brown}5\; 1\quad \quad \end{align}$$
The color-coding and $\LaTeX$ transcription is mine in the quote, trying to reflect my problem understanding this.
The actual formulation in the quote is:
$$\{\{\{1,2\},\{0,0\} \},\{\{3,0\},\{4,5\}\}\}$$
represents
$$e_1 \wedge e_2 \wedge e_3\,+\, 2\; e_1 \wedge e_3\, +\, 3\;e_2\wedge e_3\, +\, 4\; e_3\, +\, 5 $$
For instance, $\{\color{blue}1,\color{red}2\}$ (iff the color-coding makes sense) combines the coefficient $1$ of $e_1\wedge e_2\wedge e_3,$ composed of the maximum combination of basis vectors from $V$ with the coefficient for a wedge of just $2$ basis vectors, i.e. $e_1 \wedge e_3.$ Likewise, I don't see, either, what the next binary element $\{0,0\}$ corresponds to.
What is the equivalence between the WL notation and the different possible wedge products, and how are the coefficients usually "stored" (presumably a regular matrix, provided they are not functions (?))?
I think they've made a mistake, writing $e_1 \wedge e_3$ when they meant $e_1 \wedge e_2$. At least, that's the only way I can make sense of the pattern.
Then the coefficients are leaves in the binary tree obtained by successively making the choices (“$e_1$ or $1$”) $\wedge$ (“$e_2$ or $1$”) $\wedge$ (“$e_3$ or $1$”). So the first number in the (flattened) list is the coefficent of $e_1 \wedge e_2 \wedge e_3$, next comes the coefficent of $e_1 \wedge e_2 \wedge 1 = e_1 \wedge e_2$, and so on, until you get to the last number which is the coefficient of $1 \wedge 1 \wedge 1 = 1$.