Current transformer energy harvesting from a mid-voltage line
A current transformer (CT) &quot;projects&quot; the secondary (burden) impedance to a very much lower impedance on the primary via the turns ratio squared. That impedance becomes parallel to the inherently low magnetization inductance that a CT will naturally have. For example: -
If the burden is 300 Ω and turns ratio is 1:100 then the burden appears as \$\frac{300}{100^2}\$ or 0.03 Ω.
So, the important thing you have to care about (in normal CT measurement applications) is that the burden impedance projected to the primary is significantly lower than the magnetization inductance impedance of the thick wire passing through the core. If you don't do this, you will get a measurement error but, in your application, you are less worried about that error.
However, any current flowing in the magnetization inductance, cannot be utilized in the secondary so, if the primary current is 10 amps, you probably want at least 9 amps to be flowing in the projected burden impedance and no more than 1 amp of it magnetizing the core.

with a core of 10^4 relative permeability

Here's where you start - you need to determine the magnetization inductance of the primary. You need to calculate this. I can make a guess that it is (say) 100 μH but that's just a guess.
However, once you have that inductance, you can calculate the reactance. I'm assuming 50 Hz and therefore 100 μH has a reactance of 31.4 mΩ. That's milliohms. So now, you want to ensure that your secondary burden projected to the primary is about 10% of that value (3.14 mΩ).
The power in the projected burden is 9 amps squared multiplied by 3.14 mΩ. That's 0.254 watts.
$$\boxed{\text{But remember I guessed that the primary inductance is 100 μH}}$$
If the magnetization inductance is 1 mH, then it has an impedance of 0.314 Ω at 50 Hz and then, your projected burden would be one tenth this value (31.4 mΩ) and take 9 amps. That's now a power of 2.54 watts. That power is realizable in the secondary circuit (with a resistance of 314 Ω).
$$\boxed{\text{It's all down to making the primary magnetization inductance as big as possible}}$$
It's got very little to do with the turns ratio. But, can you realistically make the primary inductance as high as 1 mH with 1 primary turn? Personally, I don't think it's practical but, maybe if you do the math you'll show how.

But you could really push things (at the risk of saturating the core) and assume that the projected burden resistance has the same impedance magnitude as the magnetization reactance - with the same magnitude you could argue that of the ten amps flowing in the primary, 7.071 amps flowed in the projected burden resistance and, to achieve say 3 watts, you'd need a resistance of 0.06 Ω and a magnetization reactance of the same.
That turns out to be a magnetization inductance of 191 μH. Maybe that's more realizable but, the devil will always be in the detail. Also don't forget that there will be potentially more core saturation due to 7.071 amps in the magnetization inductance. You do the math.