Let's estimate the saturation magnetic field for a typical metal.
Suppose that permanently-magnetized iron has one freely-orientable electron for every atom.
(Reality is more complicated than this; it's a toy model.)
The magnetic moment of each electron is roughly $\mu_B = 9.3\times10^{-24}\rm\,J/T \approx 10^{-23}\rm\,J/T$.
Given iron's density around $8\rm\,g/cm^3$ and atomic weight $56\rm\,g/mol$,
each atom occupies a volume of about $12\times10^{-30}\rm\,m^3$.
The magnetic dipole moment is the volume integral of the magnetization,
$$
\mathbf m = \int \mathbf M \,\mathrm dV
$$
and the magnetization $\bf M$ is proportional to the residual magnetic field from the relation
$$
\mathbf B = \mu_0(\mathbf M + \mathbf H).
$$
So we can (in a hand-waving way) predict the saturation field of our iron as
\begin{align}
\mathbf B &= \frac{\mu_0 \mathbf m_\text{electron}}{V_\text{atom}}
\\ &= 4\pi\times10^{-7}\frac{\rm H}{\rm m} \cdot \frac{10^{-23}\rm\,A\,m^2
}{12\times10^{-30}\rm\,m^3}
\\ &\approx 1\,\rm T
\end{align}
So in this toy model, a perfect crystal of completely-magnetized iron would have a residual field of about one tesla.
You can imagine modifying the model a bit to boost this maximum. Iron has a partially-filled $d$-shell with two unpaired electrons, which would naively double this maximum field to $2\rm\,T$. And you can make a handwaving argument about g-factors
and orbital angular momentum:
since each valence electron has $L=2$ units of orbital angular momentum,
you might be able to construct iron atoms where there are more $d$-shell electrons orbiting clockwise than counterclockwise. However that doesn't appear to be how real iron works --- my reading suggests that the various $d$-orbitals aren't very well-localized in the isotropic iron crystal, but I'm not enough of a materials person to articulate why.
If you used a lanthanide or actinide metal, where the valence electrons are in the $f$-shell rather than the $d$-shell, you could have additional unpaired electrons, and orbital angular momentum could be more important. And the two strongest available magnet alloys are made with neodymium and samarium.
However the actual
neodymium alloy,
$\rm Nd_2 Fe_{14} B_2$,
is still mostly iron.
So as an order-of-magnitude guess, our maximum --- something like $2\rm\,T$ remnant field, based on approximately two spin-oriented electrons per atom --- is still appropriate.
A material with a remnant magnetization of $10\rm\,T$ isn't crazy, but it would have a very different electronic structure than the magnetic materials that we have now.
