Definition: An $n$-field is a structure $(S,\times_0,\times_1,\times_2,\cdots,\times_{n-1},e_0,e_1,e_2,\cdots,e_{n-1})$, where the set $\{e_0,e_1,e_2,\cdots,e_{n-1}\}\subseteq S$ has size $n$ (so all the $e_k$ are distinct), and each operation $\times_k$ is commutative, is associative, has identity $e_k$, is invertible on $S\setminus\{e_0,e_1,e_2,\cdots,e_{k-1}\}$, and (if $k\neq0$) distributes over $\times_{k-1}$.
I'm not sure if this is the "right definition", but that's a different question.
A $0$-field is just a set $(S)$. A $1$-field is an abelian group $(S,+,0)$. A $2$-field is a field $(S,+,\times,0,1)$.
A $3$-field $(S,+,\times,\wedge,0,1,e)$ has the following properties: For all $a,b,c\in S$,
$$a+b=b+a,\quad a\times b=b\times a,\quad a\wedge b=b\wedge a,$$
$$(a+b)+c=a+(b+c),\quad(a\times b)\times c=a\times(b\times c),\quad(a\wedge b)\wedge c=a\wedge(b\wedge c),$$
$$0+a=a,\quad1\times a=a,\quad e\wedge a=a,$$
$$\exists(-a):a+(-a)=0,$$
$$a\neq0\implies\exists(a^{-1}):a\times(a^{-1})=1,$$
$$0\neq a\neq1\implies\exists(a^\vee):a\wedge(a^\vee)=e,$$
$$a\times(b+c)=(a\times b)+(a\times c),\quad a\wedge(b\times c)=(a\wedge b)\times(a\wedge c),$$
$$0\neq1\neq e\neq0.$$
Here are some consequences of the definition: For any $a$ other than $0$ or $1$, we have
$$a\wedge(e\times0)=(a\wedge e)\times(a\wedge0)$$
$$a\wedge(0)=(a)\times(a\wedge0).$$
If $a\wedge0\neq0$, then we divide by that to get $1=a$, a contradiction. Thus $a\wedge0=0$.
And for any $a$ other than $0$ or $1$,
$$a\wedge1=a\wedge(1\times1)=(a\wedge1)\times(a\wedge1),$$
which implies either $a\wedge1=0$ or $1$. If it's $0$, then we get
$$a=a\wedge e=a\wedge(e\times1)=(a\wedge e)\times(a\wedge1)=0,$$
a contradiction. Thus $a\wedge1=1$.
Distributivity also implies (using $e\wedge0=0$ and $e\wedge1=1$)
$$0\wedge0=0\wedge(e\times0)=(0\wedge e)\times(0\wedge0)=0,$$
$$0\wedge1=0\wedge(e\times e^{-1})=(0\wedge e)\times(0\wedge e^{-1})=0,$$
$$1\wedge1=(1\wedge1)\times(1\wedge e)=1\wedge(1\times e)=1\wedge(e)=1.$$
In summary, $0\wedge a=0$ for all $a$, and $1\wedge a=1$ for all $a\neq0$.
It's obvious that, if this structure is a $3$-field, then both of the sub-structures $(S,+,\times,0,1)$ and $(S\setminus\{0\},\times,\wedge,1,e)$ are fields. Conversely, if these are fields, and $0\wedge a=0$ for all $a$, then $(S,+,\times,\wedge,0,1,e)$ is a $3$-field. Perhaps an equivalent definition of an $n$-field is a chain of $n-1$ fields (thus $n$ abelian groups), where the multiplicative group of each field is the additive group of the next field, and $e_k\times_la=e_k$ for any $l>k$ and any $a\in S\setminus\{e_0,e_1,e_2,\cdots,e_{k-1}\}$.
Let's consider some examples. If the $+\times$ field has characteristic not $2$, then $-1\neq1$ but $(-1)^2=1$, which implies that the $\times\wedge$ field has characteristic $2$; in particular, for any $a$ other than $0$ or $1$, we have $(a-1)\in S\setminus\{0\}$, so
$$1=(a-1)^2=a^2-2a+1$$
$$a=2.$$
Therefore, there is only one $3$-field with $+\times$ characteristic not $2$. Explicitly:
$$\begin{array}{c|ccc}+&0&1&2\\\hline 0&0&1&2\\1&1&2&0\\2&2&0&1\end{array}\qquad\begin{array}{c|ccc}\times&0&1&2\\\hline 0&0&0&0\\1&0&1&2\\2&0&2&1\end{array}\qquad\begin{array}{c|ccc}\wedge&0&1&2\\\hline 0&0&0&0\\1&0&1&1\\2&0&1&2\end{array}$$
On the other hand, if the $+\times$ field has characteristic $2$ and is finite, with size $|S|=2^m$, then $|S\setminus\{0\}|=2^m-1$ must also be a power of a prime $p^k$, since the $\times\wedge$ structure is also a finite field. Clearly $p$ must be odd. The multiplicative group of $\mathbb F_{2^m}$ is cyclic (it has a single generator), but the additive group of $\mathbb F_{p^k}$ is not cyclic unless $k=1$. So we must have $2^m-1=p$ where $p$ is a prime number; the $+\times$ field is $\mathbb F_{p+1}$ and the $\times\wedge$ field is $\mathbb F_p$.
If the $+\times$ structure is the infinite field of rational expressions over $\mathbb F_2$, then any element $r(x)\neq0(x)$ can be factored in terms of the irreducible polynomials:
$$r(x)=x^{n_1}(x+1)^{n_2}(x^2+x+1)^{n_3}(x^3+x+1)^{n_4}(x^3+x^2+1)^{n_5}\cdots,$$
where $n_k$ are integers. This shows that the $\times$ group is isomorphic to $\mathbb Z^\mathbb N\cap c_{00}$ (the integer sequences with finite support), which cannot be made into a field.
(To prove that it cannot: Suppose it has a multiplicative identity $\vec e\neq\vec0$, and write $\vec e=d\vec u$, where $d=\text{GCD}(\vec e)$ and $\vec u$ is an integer sequence with $\text{GCD}(\vec u)=1$. We have $\vec e^2=\vec e$, so $d^2\,\text{GCD}(\vec u^2)=d\,\text{GCD}(\vec u)=d$; that is, $d\,\text{GCD}(\vec u^2)=1$, which implies $d=1$ since these are integers. Now $2\vec e\neq0$ is supposed to have an inverse $\vec x$; but $2\vec e\,\vec x=\vec e$ implies $2\,\text{GCD}(\vec e\vec x)=1$, which is absurd; $1$ is not an even number.)
Is there an infinite $3$-field? Is the multiplicative group of some infinite field (with characteristic $2$) isomorphic to the additive group of some field?
The latter, the $\times\wedge$ field, must have characteristic $0$, since otherwise $a^p=1$ for all $a\in S\setminus\{0\}$, but such a polynomial equation has only finitely many solutions in any field. Thus, the $\times$ group must in fact be a vector space over $\mathbb Q$.
There is only one $4$-field $(S,\times_0,\times_1,\times_2,\times_3,e_0,e_1,e_2,e_3)$ where both of the structures $\times_0\!\!\times_1\!\!\times_2$ and $\times_1\!\!\times_2\!\!\times_3$ are $3$-fields. If the $\times_1\!\times_2$ field has characteristic $2$, then the $\times_0\!\!\times_1\!\!\times_2$ structure must be the $3$-field $\mathbb F_3$ described above, but then $|S|=3$, while a $4$-field is supposed to have size at least $4$. Therefore, the $\times_1\!\times_2$ field must have characteristic not $2$, so the $\times_1\!\!\times_2\!\!\times_3$ $3$-field is $\mathbb F_3$, and $|S|=4$.
$$\begin{array}{c|cccc}\times_0&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_1&e_2&e_3\\e_1&e_1&e_0&e_3&e_2\\e_2&e_2&e_3&e_0&e_1\\e_3&e_3&e_2&e_1&e_0\end{array}\qquad\begin{array}{c|cccc}\times_1&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&e_1&e_2&e_3\\e_2&e_0&e_2&e_3&e_1\\e_3&e_0&e_3&e_1&e_2\end{array}$$
$$\begin{array}{c|cccc}\times_2&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&e_1&e_1&e_1\\e_2&e_0&e_1&e_2&e_3\\e_3&e_0&e_1&e_3&e_2\end{array}\qquad\begin{array}{c|cccc}\times_3&e_0&e_1&e_2&e_3\\\hline e_0&e_0&e_0&e_0&e_0\\e_1&e_0&e_1&e_1&e_1\\e_2&e_0&e_1&e_2&e_2\\e_3&e_0&e_1&e_2&e_3\end{array}$$
But there may be $4$-fields without the property of $\times_1\!\!\times_2\!\!\times_3$ being a $3$-field.
In any case, it follows that there is no $5$-field with $\times_1\!\!\times_2\!\!\times_3$ being a $3$-field, since $S$ would have to have size $4$ but also at least $5$.
