I think you're right (for all positive integer powers).  Here's my reasoning:

The assumption that R is an escape radius means \[ z_m > R \ge c \]
implies \[ |z_{m+1}| = |z_m^n + c| > |z_m|^n - |c| > R^n - R = R (R^{n-1} - 1) \]

Taking the limit as \( |z_m| \to R \) from above gives
\[ \frac{|z_{m+1}|}{|z_m|} > R^{n-1} - 1 \ge 1 \]
=>
\[ R^{n-1} \ge 2 \]
=>
\[ R \ge \sqrt[n-1]{2} \]

which is what you claimed.  The existence of a c in the set with |c|=R is proof of minimality for some powers, but the tips where \( z_3 = z_2 \) for \( z \to z^3 + c \) have magnitude \( \sqrt[4]{3} \) which is a little bit less (at least, I think they are the relevant tips..).  But I think the smallest escape radius just has to be >= the largest c, equality is not guaranteed...

EDIT I suggest checking R=1.36 (which is between the two escape values I calculated) for the cubic Mandelbrot and inspecting carefully for any problems... maybe I'm talking nonsense...

EDIT2 these are not the cubic tips you are looking for

If you want a tighter restriction, keep in mind that the minimum radius varies with |c|. Doing that will give images where the iteration contours come much closer to the tips of the set. I implemented this in Ultra Fractal several years ago in the "elliptical bailout" formulas.

These are not Mandelbrot sets, but I do know that for fractals such as the Burning Ship, Buffalo, Perpendicular, etc. the western spike of the Nth order fractal ends at -(2^(1/(n-1))). For SOME fractals (e.g. the Mandelbrot, Quasi Heart), it appears to be that the fractal is always bounded in the circle with that radius. I might experiment around soon and see which fractals break the bounding rules.

Even though \( \sqrt[n-1]{2} \) is maybe not the lowest escape radius for the odd powers, it still has a beautiful property: it's the smallest such that the iteration bands are connected. The first iteration band is the only one that is infinitely thin anywhere, and it is so at n-1 points. With a larger escape radius, iteration bands are not infinitely thin anywhere.

I lowered the escape radius to 1.4 (slightly less than \( \sqrt{2} \)) for the 3rd power Mandelbrot set and you can see in the image that the iteration bands are not connected anymore.

With the escape radius set to \( \sqrt{2} \), the first and the second iteration bands' boundaries touch because:
After 1 iteration, z_1 = c, so points c that escape after 1 iteration have abs(c) > escape_radius.
After 2 iterations, z_2 = c^3 + c, so points c that escape after 2 iterations have abs(c^3 + c) > escape_radius.
Now fill in \( \sqrt{2}*i \) for c in both cases (this c is the "highest"/most imaginary/most northern value of c that is not larger than the escape radius).
\( abs(\sqrt{2}*i) = \sqrt(2) \), so this c is at the boundary of the first iteration band.
\( abs((\sqrt{2}*i)^3 + \sqrt{2}*i) = abs(-2*(\sqrt{2}*i) + (\sqrt{2}*i)) = \sqrt(2) \) as well, so it's also at the boundary of the second iteration band.

After 3 iterations the expression turns into \( (c^3 + c)^3 + c = (-\sqrt(2)*i)^3 + \sqrt(2)*i = 3*\sqrt(2)*i \), which is larger than the escape radius, so only the first and second iteration bands' boundaries touch.

With even powers, the iteration bands all need to go around the spikes and are infinitely thin there, so they're all infinitely thin at n-1 points (the number of spikes).

The connectedness of iteration bands is something that I'm used to and I like it. I also wanted to know the lowest escape radii mainly because I didn't like an arbitrary choice. I was surprised that no fractal programs had a good choice built in. Choosing the escape radius to be the lowest so that iteration bands are still connected is a good justification for the choice in my opinion so I can live with that. it turns out the thing I asked for was not what I was looking for because of the unexpected fact that the best escape radii for odd powers are not the lowest like for even powers.

It would still be interesting to know the lowest possible values. Because of the crazy shape of the branches of the 3rd power Mandelbrot set I'm afraid it's not even a nice expression but an infinite sum or something.

I think the problem of minimal escape radius is equivalent to finding the bounding circle of the anti-Buddhabrot for the formula?

I don't know what that is.

The difference is bigger with higher powers, perhaps obviously.  Here's power 5 Mandelbrot:

high bailout


bailout=2


low bailout (new feature in next KF, not yet released)