While related to laptop or smartphone displays, it's lighting related!
Modern quality displays go up to about 1,000 nits outdoors. Some go even above that while displaying HDR stuff (1,200 nits - iPhone 13).
What about light sources?

Let's take an F40T12 like Philips Alto with 2,500 lumens.
It's 1.5" diameter (38.1 mm), 48" long. Let's say its lighted part is 46.5" (1181.1 mm).

Now it's getting a bit difficult. What's the apex angle? Couldn't find anything but the FT light is pretty omni-directional. Let's say 300 degrees...? (I know the emission isn't totally uniform or homogenous...)
The three-dimensional angular span for an apex angle, using Ω for the angular span (in steradian) and 2θ for the apex angle, is (souce):
Ω = 2π(1 − cos 2θ/2) = 2π(1 − cos 150) = 11.724 sr

Candela to lumens calculation is (source):
Φv(lm) = Iv(cd) × Ω(sr), so Iv = Φv / Ω
Our F40T12 is 2500 lm / 11.724 sr = 213.23 cd (using the number reversely in the calculator + our apex angle I get 2500 lm so it seems ok)

What about nits? These are just candela per square metre (wikipedia).
The surface area of our cylindrical tube in metres is L = 2πrh = 2π * 0.0381/2 * 1.1811 = 0.141 m2 (Wikipedia).
213.23 cd / 0.141 m2 = 1511 nits

So our F40T12 is 1.5x brighter than iPhone 13 at the maximum brightness outdoors, which should be 1000 nits (source).
Now you can compare and tell if my calculations were correct.

Interesting calculations!  I believe there may be a slightly easier route to arrive at a similar result.

1 candle power (CP) = 4 pi lumens = 12.568 lm.  So 2500 lm = 198.82 CP
I think 1 candela (cd) is equivalent to 1 candle power per m².

The lighted surface of an F40T12 according to your calculation is 0.141 m²
So we have a surface luminance of 198.82 / 0.141 = 1411 cd/m² or nits.

Your calculation would be better if you used a diameter of 36.9mm instead of 38.1mm - the light-generating surface is of course on the inside rather than the outside of the T12 glass, and the wall thickness is usually around 0.6mm for the more common fluorescent tubes made on high speed horizontal machinery.

Thanks, James, for the simplified calculation.
It's the same result as with my calculation assuming our linear FL 100% omni-directional (360°), i.e. Ω = 4π sr

The glass thickness is a good point! Didn't realise that.
Corrected lighted surface is 2π * 0.0369/2 * 1.1811 = 0.1369m² then and it's brightness 198.82 / 0.1369 = 1452 nits.