July 2020 Edit:
Let me give more details on the relation between the above general methodology and the particular FKT construction.
First some notation: I will write $\mathbb{N}=\{0,1,2,\ldots\}$, and I will denote the set functions from the set $X$ to the set $Y$ by $\mathscr{F}(X,Y)$.
We start from the $\ell^1$ space $E$ defined as the set of functions $f\in\mathscr{F}(\mathbb{N},\mathbb{C})$ such that
$$
||f||_E:=\sum_{i\in\mathbb{N}}|f(i)|
$$
is finite.
The first step is to understand the algebraic tensor product $E\otimes E$. The general construction proceeds via the free vector space with basis indexed by symbols $f\otimes g$ with $f,g\in E$ and quotienting by relations $(f_1+f_2)\otimes g-f_1\otimes g-f_2\otimes g$ etc. Another equally uninspiring construction is to take an uncountable Hamel basis $(e_i)_{i\in I}$, for $E$, produced by the Axiom of Choice, and realize $E\otimes E$ as the subset of $\mathscr{F}(I\times I,\mathbb{C})$ made of functions of finite support (equal to zero except for finitely many elements of $I\times I$). The proper definition is as a solution to a universal problem: $E\otimes E$ together with a bilinear map $\otimes:E\times E\rightarrow E\otimes E$ must be such that for every vector space $V$ and bilinear map $B:E\times E\rightarrow V$, there should exist a unique linear map $\varphi:E\otimes E\rightarrow V$ such that $B=\varphi\circ\otimes$. One can construct such a space more concretely as follows.
Let $E_2$ be the subset of $\mathscr{F}(\mathbb{N}^2,\mathbb{C})$ made of functions $h:(i,j)\mapsto h(i,j)$ which are finite sums of functions of the form $f\otimes g$ with $f,g\in E$. Here $f\otimes g$ is the function $\mathbb{N}^2\rightarrow\mathbb{C}$ defined by
$$
(f\otimes g)(i,j)=f(i)g(j)
$$
for all $i,j\in \mathbb{N}$. Note that the definition I just gave also provides us with a bilinear map $\otimes:E\times E\rightarrow E_2$.
Proposition 1: The algebraic tensor product of $E$ with itself can be identified with $E_2$.
The proof relies on the following lemmas.
Lemma 1: For $p,q\ge 1$, suppose $e_1,\ldots,e_p$ are linearly independent elements in $E$ and suppose $f_1,\ldots,f_q$ are also linearly independent elements in $E$. Then the $pq$ elements $e_a\otimes f_b$ are linearly independent in $E_2$.
Proof: Suppose $\sum_{a,b}\lambda_{a,b}e_a\otimes f_b=0$ in $E_2$. Then $\forall i,j\in\mathbb{N}$,
$$
\sum_{a,b}\lambda_{a,b}e_a(i) f_b(j)=0\ .
$$
If one fixes $j$, then one has an equality about functions of $i$ holding identically.
The linear independence of the $e$'s implies that for all $a$,
$$
\sum_{b}\lambda_{a,b}f_b(j)=0\ .
$$
Since this holds for all $j$, and since the $f$'s are linearly independent, we get $\lambda_{a,b}=0$ for all $b$. But $a$ was arbitrary too, so $\forall a,b$, $\lambda_{a,b}=0$ and we are done.
Lemma 2: Let $B$ be a bilinear map from $E\times E$ into some vector space $V$.
Suppose $g_k,h_k$, $1\le k\le n$ are elements of $E$ satisfying
$$
\sum_{k}g_k\otimes h_k=0
$$
in $E_2$, i.e., as functions on $\mathbb{N}^2$. Then
$$
\sum_k B(g_k,h_k)=0
$$
in $V$.
Proof: This is trivial if all the $g$'s are zero or if all the $h$'s are zero. So pick a basis $e_1,\ldots,e_p$ of the linear span of the $g$'s and pick a basis $f_1,\ldots,f_q$ of the linear span of the $h$'s (no Axiom of Choice needed). We then have decompositions of the form
$$
g_k=\sum_a \alpha_{k,a}e_a
$$
and
$$
h_k=\sum_b \beta_{k,b} f_b
$$
for suitable scalars $\alpha$, $\beta$.
By hypothesis
$$
\sum_{k,a,b}\alpha_{k,a}\beta_{k,b}\ e_a\otimes f_b=0
$$
and so $\sum_k \alpha_{k,a}\beta_{k,b}$ for all $a,b$, by Lemma 1.
Hence
$$
\sum_k B(g_k,h_k)=\sum_{a,b}\left(\sum_k \alpha_{k,a}\beta_{k,b}\right) B(e_a,f_b)=0\ .
$$
Now the proof of Proposition 1 is easy. The construction of the linear map $\varphi$ proceeds as follows.
For $v=\sum_{k}g_k\otimes h_k$ in $E_2$, we let $\varphi(v)=\sum_k B(g_k,h_k)$.
This is a consistent definition because if $v$ admits another representation $v=\sum_{\ell}r_{\ell}\otimes s_{\ell}$, then
$$
\sum_k g_k\otimes h_k\ +\ \sum_{\ell}(-r_{\ell})\otimes s_{\ell}=0
$$
and Lemma 2 implies
$$
\sum_k B(g_k,h_k)=\sum_{\ell} B(r_{\ell},s_{\ell})\ .
$$
The other verifications that $E_2$ with $\otimes$ solve the universal problem for the algebraic tensor product pose no problem.
The second step is to construct a topological completion $\widehat{E}_2$ for $E_2$. I will use the projective tensor product construction $E\ \widehat{\otimes}_{\pi}E$.
For $h\in E_2$, I will use the $l^1$ norm
$$
||h||_2=\sum_{(i,j)\in\mathbb{N}^2}|h(i,j)|\ .
$$
I will also use the seminorm
$$
||h||_{\pi}=\inf\ \sum_k ||g_k||_E\times||h_k||_E
$$
where the infimum is over all finite decompositions $h=\sum_k g_k\otimes h_k$.
The projective tensor product is the completion with respect to $||\cdot||_{\pi}$.
The $||\cdot||_1$ is an example of cross norm, i.e., it satisfies
$||f\otimes g||_2=||f||_E\times||g||_E$. Moreover, one has the following easy result.
Proposition 2: For all $h\in E_2$, we have $||h||_2=||h||_{\pi}$.
For the proof use the cross norm property and triangle inequality for $\le$, and for the reverse inequality, approximate $h$ by the truncation where $h(i,j)$ is replaced by zero unless $i,j\le N$.
Now it is clear that the abstract topological tensor product $\widehat{E}_2$ is nothing but the familiar $\ell^1$ space of functions on $\mathbb{N}^2$.
Likewise (but with heavier notations) one can construct $\widehat{E}_n=E\ \widehat{\otimes}_{\pi}\cdots\widehat{\otimes}_{\pi}E$, $n$ times,
with the corresponding $\ell^1$ norm
$$
||h||_n=\sum_{(i_1,\ldots,i_n)\in\mathbb{N}^n}|h(i_1,\ldots,i_n)|\ .
$$
The topological exterior power $\widehat{E}_{n,{\rm Fermi}}$ can be identified with the closed subspace of antisymmetric functions inside $\widehat{E}_n$, namely functions $h:\mathbb{N}^n\rightarrow\mathbb{C}$ which satisfy
$$
h(i_{\sigma(1)},\ldots,i_{\sigma(n)})=\varepsilon(\sigma)\ h(i_1,\ldots,i_n)
$$
for all $(i_1,\ldots,i_n)\in\mathbb{N}^n$ and all permutations $\sigma$.
We will equip the space with restriction of the norm $||\cdot||_n$.
Now consider the algebraic direct sum $W=\oplus_{n\ge 0}\widehat{E}_{n,{\rm Fermi}}$.
Given (for the moment unspecified) positive weights $w_n$, let us define the norm
$$
||H||_{\rm Big}=\sum_{n\ge 0}w_n||h_n||_n
$$
where $H$ is an element of $W$ seen as an almost finite sequence $(h_0,h_1,\ldots)$
of functions in $\widehat{E}_{0,{\rm Fermi}},\widehat{E}_{1,{\rm Fermi}},\ldots$
Clearly the completion $\widehat{W}$ is obtained by removing the almost finite restriction but still requiring convergence of the sum defining $||\cdot||_{\rm Big}$.
Finally, to make contact with FKT, to $H=(h_0,h_1,\ldots)\in\widehat{W}$
we associate the set function $\alpha:\mathcal{J}\rightarrow\mathbb{C}$ where $\mathcal{J}$ is the set of finite subsets of $\mathbb{N}$ (including the empty set), as follows.
For $I=\{i_1,\ldots,i_n\}\in\mathcal{J}$ with $i_1<\cdots<i_n$ we let by definition
$$
\alpha(I)=h_n(i_1,\ldots,i_n)\ .
$$
If we pick the weights $w_n=\frac{1}{n!}$, then this correspondence is a bijective isometry with the giant $\ell^1$ space of FKT.
Remark: One can do the same long construction with $\ell^2$ instead of $\ell^1$ norms
and this will produce the Fermionic Fock space of the Hilbert space $\ell^2(\mathbb{N})$, as in the mathematical literature on second quantization, e.g., in the book by Reed and Simon.
Note that the corresponding topological tensor products of Hilbert spaces were introduced by Murray and von Neumann in "On rings of operators", Ann. of Math. 1936,
and further developed by Cook in "The mathematics of second quantization", PNAS 1951, for the needs of Quantum Field Theory.
