this will shift the power losses around
Increasing the supply voltage 10x will reduce the current by 10x and the (I^2 *R) loss by 100x, which is what you've shown.
But current computing equipment doesn't run on 800V, so somewhere in the chain there will need to be power converters or power supplies to get the voltage levels down to usable levels. These power converters will need to be extremely efficient to minimize that conversion loss that will also come out as heat. Go back and do some of those conversions: 15 kW at 90% efficiency means 1.5 kW is lost in the conversion process. 95% loses 750 W to heat.