Welcome. As previous answers have explained, reducing a 12V supply down to 5V by dropping 7V across a series element (in this case, the linear regulator) will result in most of the power delivered by the battery being wasted as heat. Using the numbers you provided:
Current required by the load = 800 * 60mA = 48A
Power delivered by battery = 12V * 48A = 576W
Power consumed by the load = 5V * 48A = 240W
Power lost as heat in linear regulator = 576W - 240W = 336W
Not only will this be impractical to build, but it will also drain the battery very fast. You didn't mention the type of battery, so for purposes of this discussion let's assume it is lead-acid, most lead-acid batteries are designed to be discharged at what is called "the 10-hour discharge rate", or "C10 rate", which simply means:
"C10 rate" current (amps) = Amp-Hours / 10 hours.
So a 120Ah battery has a C10 rate current (10-hour discharge rate) of:
120Ah / 10 = 12A.
As the discharge current exceeds this value, the actual charge the battery will supply reduces; meaning: the battery will deliver significantly less than 120Ah.
A 120Ah battery being discharged at 48A will not last long before needing a recharge, and its lifecycle will be reduced compared to a battery discharged at its C10 rate (12A, in this case).
Here are some suggestions:
Option #1 Single PSU
1x Switch-mode power supply (buck converter buck), rated at:
Vin=12V, Vout=5V, Pout=240W (48A @5V).
This is not a common device, but I did find one:
https://daygreen.com/en-en/products/12v-24v-to-5v-50a-250w-dc-dc-step-down-converter-voltage-regulator-1
The problem with having just one big supply is that it is quite capable of destroying (burning up) any of the 800 NeoPixels should they suffer some kind of internal fault.
Option #2 Multiple PSUs
Several PSUs of lower power, eg:
10 x Switch-mode power supply (buck converter buck), each rated at:
Vin=12V, Vout=5V, Pout=24.0W (4.8A @5V).
This is much more common and easier to source, here is just one example:
https://www.amazon.com.au/Converter-DROK-Regulator-Inverter-Transformer/dp/B01NALDSJ0?th=1
Compared with option #1, this has a far lower risk of burning up any of the loads. Also, if one of the loads fails, its effect will be limited to just that PSU (eg: that PSU may go into current-limit mode), the remaining PSUs will continue operating as normal.