Answering the question of how to get the desired field. Leaving the construction of suitable elliptic curves to you, as I have zero experience with Sage. I hope this helps in your task in some way, but I am prepared to be wrong.
This seems to be relatively close to the limit of "simple computability" in the sense that it took a bit of tinkering to coerce Mathematica to find a suitable way. I like to think Sage has been programmed to be more efficient with number theoretic crunching, so I would welcome a confirmation of all of the following by Sage or any other CAS.
My starting point is @Servaes's observation that your cardinality is $p^{12}$ with
$$
p=21888242871839275222246405745257275088696311157297823662689037894645226208583,
$$
a prime number with $77$ digits.
I want to extend the prime field $GF(p)$ by adjoining a root of unity of a smallish order $\ell$. To that end I first want to know which roots of unity are in the prime field, so I factor
$$
p-1=2\cdot3^3\cdot13\cdot29\cdot67\cdot229\cdot311\cdot983\cdot11003\cdot 405928799\cdot11465965001\cdot13427688667394608761327070753331941386769.
$$
Here the powers of $2$ and $3$ are of interest, because to get a field of cardinality $p^{12}$ we need $\ell$ to be a factor of $p^{12}-1$, but not to be a factor of $p^n-1$ for any $n$ that is a proper factor of $12$.
The well known properties of the Euler totient function suggest to look for numbers of the form $\ell=2^a\cdot 3^b$. A bit of testing reveals that
$$\ell=864=2^5\cdot3^3$$
works for us in the sense that the residue class
$$p\equiv 775\pmod{864}$$
has multiplicative order exactly twelve.
So we need to adjoin a root of unity of order $864$. To do that we need a modular factorization of the cyclotomic polynomial
$$
\Phi_{864}(x)=x^{288}-x^{144}+1
$$
in the ring $GF(p)[x]$.
I was delighted to see that the Mathematica implementation of modular factorization algorithms (don't know which they are using) worked. It took a while, but it gave a factorization of $\Phi_{864}(x)$ consisting of the expected $24$ degree $12$ factors $f_i(x)$, $1\le i\le 24$, one of which is
$$
f_1(x)=1989524725549740340342230321930199252175174949955210132320435129795479
5986100 +
469168846834798856517972474313896222613925797978349957617992393054855
137997 x^6 + x^{12}.
$$
Furthermore, all the factors $f_i(x)$ have the shape $f_i(x)=g_i(x^6)$, where the polynomials $g_i(x)$ are irreducible quadratics. It behooves me to explain this phenomenon, as it points at possible extensions of this technique (possibly widening the range of extension degrees that can be covered).
As $864=6\cdot144$, the polynomials $g_i(x)$ must be factors of the cyclotomic polynomial
$$
\Phi_{144}(x)=x^{48}-x^{24}+1.
$$
We have
$$p\equiv55\pmod{144},\qquad p^2\equiv1\pmod{144},$$
so $p$ has order two modulo $48$, whence $\Phi_{144}(x)$ factors into a product of quadratics $g_i(x), 1\le i\le 24$, modulo $p$. The catch is that if $\zeta\in K:=GF(p^2)$ is a root of unity of order $144$, then ALL the roots of the polynomial
$$h(x)=x^6-\zeta\in K[x]$$
necessarily have multiplicative order $6\cdot144=864$ because both $2$ and $3$ already appear as factors of $144$. So any zero of $h(x)$ generates the desired
field $L:=GF(p^{12})$. As $[L:K]=6$, it follows that $h(x)$ is irreducible over $K$ and that $g_i(x^6)$ must also be irreducible over the prime field.
Anyway, you can construct $L$ as
$$
L=GF(p)[x]/\langle f_1(x)\rangle
$$
in the usual way.
GF(). – user2284570 Jul 16 '24 at 20:16GF(-)doesn't seem to work with it, at least not in a reasonable amount of time. – John Palmieri Jul 16 '24 at 20:49GF(5**32)works fine on my machine, returning almost immediately, butGF(p**6)does not, wherepis the prime number you're using. A number like $5^{32}$ is much smaller than the numbers you're using. – John Palmieri Jul 16 '24 at 23:14