Characteristic 3 analogue of nimbers?

Importantly, the natural numbers with $\leq 2^n$ binary digits already form a field $F_n$ of order $2^{2^n}$ with the nim operations.

Let's see how this works inductively. $F_n \subset F_{n+1}$ is a degree $2$ extension. We have the (nim) identity $(2^{2^n})^2 = 2^{2^n} + 2^{2^n-1}$. In other words, $2^{2^n}$ is a root of the irreducible polynomial $T^2 - T - 2^{2^{n}-1}$ over $F_n$ (note: it is irreducible over $F_n$ precisely because $2^{2^{n}-1} \in F_n \setminus F_{n-1}$).

This gives a much simpler description of the $2$-nimbers: they're just the usual tower of finite fields of order $2^{2^0}, 2^{2^1},$ etc., but at stage $n \to n+1$ we identify $2^{n+1}$ with a root of $T^2 - T - 2^{2^{n}-1}$.

To guarantee uniqueness, we need to assume two things: Addition-without-carrying, and the law that the nim-product of $2^{2^k}$ with $2^{2^l}$ is the ordinary product of these when $k\neq l$. This gives a unique multiplication on the binary numbers with $\leq 2^{n+1}$ digits, that is compatible with the previous one.

It appears that every step of this applies equally to the "$p$-nimbers": Start with $F_0 = \mathbb{Z}/p$, identified as the naturals with one base $p$ digit, and form the extension $F_n \subset F_{n+1}$ using addition without carry, identifying $p^{n+1}$ with a root of the irreducible polynomial $T^p - T - p^{p^n - 1}$, and transporting the existing field structure from the finite field of $p^{n+1}$ elements.

To ensure uniqueness, we enforce that the nim-product of $p^{p^k}$ with $p^{p^l}$ is their usual product. This works because every power of $p$ is a product of powers of this form, and using the same idea we can show that the induced multiplication is well-defined and satisfies the right properties.

This makes $\mathbb{N}$ into a field of characteristic $p$, and we can all feel very pleased with ourselves.

You should see the paper "On On_p", by Joseph DiMuro. In it, he gives a characteristic $p$ analogue of the nimbers. The addition is given by adding base $p$ without carries (including for ordinals). The definition of multiplication, unfortunately, is not so simple. But I think it's a good generalization.

The definition for the natural numbers is almost identical to the one Slade gives above; the only difference is in the choice of irreducible polynomial. This will result in a different field structure -- DiMuro's is equal to the union of $\mathbb{F}_{p^{2^k}}$, while (if I'm understanding correctly) Slade's is equal to the union of $\mathbb{F}_{p^{p^k}}$. But it's clearly very similar.