I don't think there is anything built in, but here is a naive implementation -- trial division without any redundancy checks.
Note this only will factor over Euclidean domains of the form $\mathbb{Z}[\alpha]$.
(* FactorIntegerED[{a, b}, ω, Nrm] factors a + b ω over Z[ω] with norm Nrm *)
FactorIntegerED[{m_, n_}, ω_, norm_] := Module[{a = m, b = n, div, res, cands, x, y, unit},
div = Select[Divisors[norm[m, n]], PrimeQ];
res = Reap[
Do[
cands = Reverse[{x, y} /. Solve[norm[x, y] == div[[i]], {x, y}, Integers]];
Do[
While[divides[{a, b}, cands[[j]], ω],
Sow[cands[[j]].{1, ω}];
{a, b} = divide[{a, b}, cands[[j]], ω]
],
{j, Length[cands]}
],
{i, Length[div]}
]
][[-1]];
If[Length[res] == 0 || Length[First[res]] == 1,
{{m + n ω, 1}},
res = Tally[First[res]];
unit = Solve[(Times@@Power@@@res)(q + w ω) == m + n ω && {q, w} ∈ Integers, {q, w}];
Prepend[res, {First[{q, w} /. unit].{1, ω}, 1}]
]
]
divide[{a_, b_}, {x_, y_}, ω_] := divide[{a, b}, {x, y}, ω] =
With[{sol = Solve[q + w ω == (a + b ω)/(x + y ω) && {q, w} ∈ Integers, {q, w}]},
If[!MatchQ[sol, {{_, _}}],
$Failed,
First[{q, w} /. sol]
]
]
divides[e__] := ListQ[divide[e]]
This is very slow but gets the job done (for you example at least, I haven't tested if out for anything else).
FactorIntegerED[{7, 0}, (-1)^(2/3), Function[{a, b}, a^2-a b+b^2]]//AbsoluteTiming
(* {1.840636, {{-(-1)^(2/3), 1}, {3+2 (-1)^(2/3), 1}, {3+(-1)^(2/3), 1}}} *)
Edit: We can use this slow code to see which prime integers are still prime over $\mathbb{Z}[\omega]$.
myCenterDot[e_] := e
myCenterDot[l__] := CenterDot[l]
mySuperScript[p_, e_] := HoldForm[p^e]
ω = (-1)^(2/3);
Nrm = Function[{a, b}, a^2 - a b + b^2];
P = Prime[Range[26]];
Grid[Transpose@Partition[
# == myCenterDot @@ mySuperScript @@@ FactorIntegerUFD[{#, 0}, ω, Nrm] & /@ P,
13], Spacings -> {4, .75}, Alignment -> Left] /. ω -> "ω" // TraditionalForm

Luckily Wikipedia agrees with this output.