Office equipment doesn't really have "transformers" anymore, everything uses switched-mode power supplies. The way they work, is that they have an output buffer (a capacitor or inductor) and rapidly turn a switch on and off to keep that output buffer at the right level: if the output buffer is below the target voltage, the switch closes and the voltage will start to rise. Once the output buffer rises above the target voltage, the switch opens and the voltage will drop as power is being used from the buffer. This happens thousands or even millions of times per second. This principle works exactly the same for modern AC/DC power supplies as it does for DC/DC ones.

The problem with your plan is Ohm's Law, and more specifically that fact that wires aren't perfect and have some resistance (for now!). Ohm's Law gives us V=I*R, where V is the voltage in volts (V), I is the current in ampere (A), and R is the resistance in Ohm (Ω). In a wire, the resistance is constant, and V is the voltage loss across the wire. So how do we reduce the loss? We reduce I. Luckily we only actually care about the total power, which is given by P=U*I. If we want the power to stay the same and reduce the current, we have to increase the voltage.

Let's say the two wires from our central transformer to the computer are 14-gage copper, and they are 100 feet long. Their resistance is about 0.5Ω combined. We want to power a 120W computer. If we transfer that at 12V (the normal voltage computers use internally), we'd have to transfer 120W/12V=10A. The voltage loss across our wires is 10A*0.5Ω = 5V! So we put in 12V, but get out only 7V as we burned 50W in the cable itself. To get out the desired 12V we'd have to put in 17V at 10A instead, or 170W to power a 120W computer. It would also mean supplying way too high of a voltage to a computer connected with a 3-foot cable.

If we increase the voltage across the wires to 120V and down-convert that to 12V at the computer we'd only need to conduct 1A and the wire loss would be 0.5V, which at 1A is a power loss of 0.5W. That's completely acceptable, and because the computer down-converts anyways we don't really have to care about it getting 119.5V instead of 120V either.

But now we are back with a power supply at each individual computer, so in the end we didn't really gain anything. Instead of an AC/DC power supply in every computer we now have a virtually identical DC/DC power supply, so what's the point? You might have some small gains by doing the initial AC/DC conversion centrally, but in practice it probably isn't enough to care. It is only really worth it when your power comes from DC anyways, like an office with rooftop solar.

Alternatively we can use way thicker cables, but to get that same 0.5W loss at 10A would mean a wire with a resistance of 0.005Ω. To illustrate, that means using two 0000 AWG wires in parallel.