Nocturnal Ice Making: is it possible and by what mechanism? It is claimed by Wikipedia that:

Nocturnal ice making in early India and Iran
In India before the invention of artificial refrigeration technology,
ice making by nocturnal cooling was common. The apparatus consisted of
a shallow ceramic tray with a thin layer of water, placed outdoors
with a clear exposure to the night sky. The bottom and sides were
insulated with a thick layer of hay. On a clear night the water would
lose heat by radiation upwards. Provided the air was calm and not too
far above freezing, heat gain from the surrounding air by convection
was low enough to allow the water to freeze. A similar technique
was used in Iran.

So the set up would something like this:

Where both $T_a$ and $T_w$ are close to $0^{\circ}\mathrm{C}$ but $T_a > T_w$.
We're then lead to believe that due to radiative losses the water will freeze (turn to ice).
In another version (with the same set up) it is claimed the mechanism for cooling and subsequent freezing is evaporative cooling of the water.
This too I find somewhat incredible because $T_a>T_w$.
So is this method of freezing water effectively possible and if so, by what mechanism?
 A: Absolutely.  On a clear, windless, dry night, get one of those IR thermometers and point it up at the sky.  Temperatures below -40C can be seen in lots of places if it's dry enough.  The less water vapor in the air, the cooler it can appear.
The thermal environment for the water is:

*

*conduction with the tray (chosen for low conductivity)

*conduction with the air (low conductivity)

*convection with the air (very low on a still evening with the water cooler than the air)

*radiation above the water (40 degrees cooler not uncommon)

Together, these allow the water to equilibrate at a temperature below freezing, even if the air is above freezing.  Radiative transfer at these temperatures isn't very fast, so it will take a few hours.
A: Possible but requires very specific circumstances.
Suppose we put the same bowl of water on the outside of a space satellite orbiting in the thermosphere and position huge mirrors in nearby orbit to block the radiation of the sun. The near-vacuum atmosphere there is at temperatures above 2000 K (1). The water will never equilibrate to 2000K. The thermosphere will always be heating the bowl and the bowl will always be cooling the thermosphere, but because the density of the bowl is millions of times higher, it will lose heat by radiation far faster than it gains heat by thermal contact. The bowl will eventually equilibrate to a tiny fraction of a Kelvin above the radiation temperature of the surrounding space, about 200K if the bowl is illuminated by Earth's thermal radiation, or about 3K if it has to make do with the radiation of the distant stars.
So radiative cooling can, under the correct circumstances, allow water to freeze. Can those circumstances exist on Earth? Yes, for air temperatures very close to freezing, and for very still air and very good insulation slightly above freezing, since air's thermal conductivity is very low.
In the limit as $T_{W_i} \to T_{A_i}$, $dQ_{thermal} \to 0$ (assuming negligible contribution from the ground through the insulation.
$$\frac {dQ_{radiation}}{dt} \approx σεA(T_W^4 -T_{atm}^4) $$
where σ is the Stefan-Boltzmann constant, A is the surface area exposed to the sky, and $T_W$ is its temperature in kelvin, and $T_{atm}$ is the ambient temperature due to radiation of the night sky at Earth's surface. ε is the emissivity of the surface, a unitless number between 0 (a mirror) and 1 (a perfectly black surface). Water is about 0.96, depending on the wavelength under consideration.
https://physics.stackexchange.com/a/153947/285671 has a discussion of how to approximate the temperature of the night sky from below.
Approximate $T_W = 273.3K \pm (T_{W_i}-273.3K)$ and retain error instead of making the calculus complicated, since our calculation isn't going to have more than two significant figures anyway.
Integrate over, say, six hours from the last haze of sunlight in the west to the first predawn glow in the east to get $Q \leq \Delta t \frac {dQ}{dt}$. (It's less because the emissivity of the water will fall as it begins to freeze.) Reduce Q by $4.2\frac {J}{gK}$ until you reach 273.3K and then divide what's left by 333.55 J/g and your answer is in grams of ice.
1: Direct impacts of solar rays with air molecules (esp. oxygen) allows the temperature (that is to say average kinetic energy of the molecules) to far exceed the effective radiation temperature of the sun at earth's distance, some 393 K.

Regarding evaporative cooling, that won't contribute much to freezing, but it will absolutely cool the water below the ambient air temperature. You can do this experiment yourself. Get a water bottle at room temperature. Put a few thick socks on it. Soak the socks in more room temperature water. Suspend the bottle in your shower and wait an hour. Measure the temperature of the water in the bottle. You won't find any ice in the water, but it will be colder than the air temperature. The effect will be more noticeable the hotter and drier the air.

edit 2/13/21 - replaced guess of sky radiation temperature with link to a formula for predicting it
