Between $1/40,000$ and $1/8,000,000$ of the brightness as seen from Earth, depending on what the actual orbit would turn out to be, and where the planet is in its $15,000$ year orbit period.
Brightness drops as $dfrac{1}{r^2}$ with distance from the light source. Earth is at $1~textrm{AU}.$ The theoretical planet is at $200~textrm{AU}$ when it's closest to the sun, and up to $2800~textrm{AU}$ at the point furthest from the sun at the upper end of the estimated orbit.
So e.g. $1/200^2 = 1/40,000$ of the brightness (luminance) as seen from Earth.
For how a human would experience it, we can convert to exposure value as used in photography: The difference in exposure value (photographic 'stops') is $log 2$ of the luminance ratio, so we would have $15$ to $23$ stops less light than on Earth.
Sunny noon on Earth is $15~textrm{EV}.$
So the brightness at noon on the planet surface would be:
At $200~textrm{AU},$ planet orbit is closest to the sun: About $0~textrm{EV},$ roughly the same as a dimly lit interior
$400~textrm{AU},$ lower bound on semi-major axis: $-2 ~textrm{EV},$ similar to a landscape lit by the full moon
$1500~textrm{AU},$ upper bound on semi-major axis: $-6~textrm{EV},$ similar to landscape lit by a quarter moon
$2800~textrm{AU},$ upper bound on aphelion (the point on orbit most distant from the sun): $-8~textrm{EV}.$ This would be dark, but you would probably still see enough to avoid running into things.
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