Density of air at sea level: 1225 g/m^3
C02: 0.0383% by volume (383 ppmv) corresponds to 0.0582% by weight.
Ergo, 1m^3 of air has 0.713 g of C02 in it.
Ergo pulling 1 metric ton (10^6 grams) of C02 per day requires processing AT LEAST : 10^6/ .713 = 1.4m m^3 of air per day or 16 m^3 of air per second!
(That would assume 100% capture)
I was skeptical that 1 cooling tower generates this much flow, but the example in [1] suggests 17*10^6 ft^3/minute, or roughly 8000 m^3/sec.
Thus, as long as your capture chemical has 2% efficiency, it seems reasonable.
[1] https://www.power-eng.com/emissions/cooling-tower-heat-trans...
[EDITED after I detected an error in my math]
The amazing observation for me is that evaporative cooling towers process A LOT OF AIR per second.