Is the downward pressure exerted by raindrops in even a strong storm on a lightweight aircraft negligible?
Someone better informed may likely improve the following reasoning, cobbled together from haphazard sources.
Heavy rain is 3 inches per hour, or 0.021 mm/s of accumulation. (Maybe a lot more while flying through a hurricane, but I couldn't find hard numbers.)
Per m$^2$ of impacted horizontal surface, that's 21000 mm$^3$ / s / m$^2$, or 21 cm$^3$…, or 21 g/s/m$^2$.
All else being equal, faster falling raindrops should exert more pressure.
Now use a few facts from the paper "Experimental determination of forces applied by liquid water drops…" (Yu & Hopkins, 2018), which models and measures raindrops falling onto smooth glass, similar to an airplane's skin.
From fig. 3, a fast terminal velocity of a raindrop is 9 m/s, for a (largish) 4.5 mm raindrop. (Some argue that larger drops break up; others, that they are squashed rather than spherical; but this recent paper seems trustworthy.)
For such a drop at such a speed, the first row of table 3 gives values of C$_1$=2.7186, $α_1$=3.3367, $β_1$=1.5027 to plug into eq. 33, the raindrop's force $F$ (N) as a function of elapsed time $t$ (s) since impact:
$F(t) = C_1$ exp(−(ln $1000t$ +$α_1$)$^2$ / $β_1^2$) N
$F(t) = 2.7186$ exp(−(ln $1000t$ + 3.3367)$^2$ / 2.2581) N.
Integrating that from $t=0$ to $0.004$ s, the duration over which $F(t)$ is non-neglible, yields an impulse of 0.0004526 N$\cdot$s/drop.
Each 4.5 mm spherical raindrop is 0.0477 cm$^3$, thus 0.0477 g.
So at 21 g/s, that's 21/0.0477 = 440 raindrops per second.
So the overall force is 440 drop/s $\cdot$ 0.0004526 N$\cdot$s/drop = 0.2 N.
Per m$^2$, that's a pressure of 0.2 N/m$^2$.
The airplanes with the lightest wing loading are paragliders, at about 3 kg/m$^2$. In terrestrial gravity, that's about 30 N/m$^2$.
So flying one of those through a heavy downpour would require some extra lift, but only 1/150th as much as the wing is already making. A Boeing 747, at 650 kg/m$^2$, would need only 1/30000th extra.
John Hunter's answer suggests that the weight of water buildup matters more than the pressure of impinging raindrops. I found confirmation in a 1982 NASA report:
For 500 mm/h rain (6 times more than my guessed maximum),
an airliner gets 2600 N of extra downforce (and 70x more drag force) (p. 23), while the water film is 0.6 to 0.8 mm thick (p. 29),
an extra weight of 2340 kg (p. 30).
So at least for this case, downward pressure is about 5 N/m$^2$, while extra weight is 4.3 kg/m$^2$ or 43 N/m$^2$, much more.
This report describes many other effects of rain on airplanes.
More recent (paywalled) papers may have measurements, not just models.
Best Answer
The impact of new drops even in heavy rain has only a small effect.
From force x time = change in momentum
$$F\times 1 = 0.021 \times 10^{-3}\times 1000\times 9$$
where the 1000 is for the density of water, the $0.2N/m^2$ is confirmed.
The weight of a thin layer of water lying on the wings may be more significant.
Even for 0.1mm, the weight is about $0.1 \times 10^{-3} \times 1000 \times 9.8$, about $1N/m^2$. So you could look into the thickness of the water on the wing.
As the water would be swept off for most aircraft, this factor would be more significant for vertical take off and landing aircraft, but still low.
To summarise, the 'downforce of rain on aircraft' seems to be negligible.