I used NASA's "Eyes on the Solar System" app to generate approximate views of what it would look like, near Europa, just before an eclipse happened:
The top view has all the disks painted in, as if the observer had a bright headlight, and the bottom view has only the sunlit parts lit up (more realistic). This app does not do atmospheric scattering, so the scattered light from the occulting body's atmosphere (in the OP's question) is not accurately shown. The Jupiter crescent (you might need to zoom in to see it) in the "natural light" view is actually sunlit atmosphere, not scattering over the limb.
The Jupiter system is ~5x farther away from the Sun than the Earth system, so the Sun appears fainter because it has a smaller angular size. These are the apparent diameters of the bodies on a random day (2022-Nov-12) according to JPL Horizons:
Earth as seen from the Moon: 1.8°
Sun as seen from the Moon: 0.5°
Jupiter as seen from Europa: 12.2°
Sun as seen from Europa: 0.1°
From our experience on the surface of the Earth, we know there is still sunlight coming over the horizon after sunset. At the Moon, that terrestrial horizon shrinks to a little ring 1.8° wide, up in the sky.
From the Moon, during an eclipse, you still get some light scattered through that little ring, but it is coming from a much smaller area. During the maximum eclipse, the Sun is only 0.9° below the Earth's "horizon" or limb, so the attenuation is not extreme, so we see the Moon as sunset-red during a total eclipse.
On Europa, the geometry is a bit different. At maximum eclipse, the Sun would be about 6° inside Jupiter's limb, and 6° is like the start of nautical twilight. So the amount of attenuation is much greater compared to the view from the Moon, and on top of that the Sun is 52 = 25 times fainter. (Also, that twilight definition is from the Sun to the observer inside the atmosphere. For an observer on the eclipsed moon, there is an additional path length back out of the atmosphere to space, giving double the attenuation.)
To summarize, the atmospheric refraction light source during eclipse is a factor of 25 fainter because of the heliocentric distance, a factor of 400 fainter because the solar distance below the horizon is greater for Jupiter than for the Earth, and a factor of 7 brighter because Jupiter's apparent size from Europa's surface is larger than the Earth's apparent size from the Moon. If this reasoning is correct, maximum solar eclipse from Europa is about 1400 times darker than maximum solar eclipse from the Moon.
The OP uses the phrase "dark as a regular night." But this varies a lot. On the Moon, a regular night with a full Earth would be very bright, maybe 30x brighter than a full-moon night on the Earth's surface. On Europa, a regular night would also vary, depending on whether Ganymede and Callisto were up or not.
This Galilean moonlight is enough to give the SRU (Stellar Reference Unit, a sensitive star-tracker) on Juno a clear view of cloud features at local midnight on Jupiter: