Each of those devices has a maximum and minimum voltage requirement. You must find the highest of all the minima, and lowest of of all the maxima, and those two values represent a voltage range that satisfies the requirements of all devices.
For example, if your SBC requires between 4.5V and 5.5V, your disk drive requires between 4.7 and 6.0V and your camera needs from 4V to 5.2V:
$$
\begin{array}{lll}
&Device &Min &Max \\
\hline
&\text{SBC} &4.5 &5.5 \\
&\text{Disk} &4.7 &6.0 \\
&\text{Camera} &4.0 &5.2 \\
\hline
&\text{Everything's happy} &4.7 &5.2 \\
\end{array}
$$
By that logic, a supply of exactly 5.0V will do the job. Of course all this assumes that they are being supplied by the same voltage source. If a device is being powered from some other source, like a USB port or external power supply, then you can't apply this logic.
Assuming you have such a 5V supply, what's of more concern to you is the current required from that 5V supply. The total current required is the sum of all the individual currents required by each device, again assuming they are all powered from that same supply. If you have an SBC rated for 2A, a disk drive requiring at most 1A, and a camera requiring up to 500mA, that's a total of 3.5A, which your supply must be able to provide.
That doesn't mean they will all be drawing this current all the time, but there is a chance that all 3.5A be needed simultaneously. So the power supply must be up to the task 100% of the time, even if the full 3.5A is needed only 0.0001% of the time. And it needs to be able to do this while staying within whatever the acceptable voltage extremes are, as defined by the devices, and as I explained above.