Electronic – the formula to convert the result of an 8-bit ADC to voltage

This isn't going to work. Between the accuracy of your resistors and the accuracy of the A/D, you're not going to get exactly one known reading for each unique switches setting. Not even close.

In general the way to produce a set of evenly spaced analog values from different switch settings is to do what most D/As do, which is to use a R-2R ladder. I have to do something else now, so don't have time to get into what a R-2R ladder is, maybe later. However, this is a well known construct and you will easily find lots about it out there.

Carefully look at your D/A error budget and that of the A/D. With .1% resistors and depending on the absolute accuracy of the A/D, you might be able to resolve 8 switches if everything works out right. 7 or 6 switches sounds a lot more doable. 10 is definitely in the guaranteed to fail range.

Ignoring the harsh reality of the performance of ADCs which can be cheaply built into MCUs, in the ideal case each output count represents a possible well-defined range of input voltages.

Here is an ideal (according to Atmel's definition) 3-bit ADC.

There are \$2^N\$ steps. In the above example, that is 8. With an ideal unipolar ADC, an input of lower than \$\frac{V_{REF}}{2^{N+1}}\$ will give an output of 0- as you can see above, below 1/16 of the 2V reference voltage the output code will be 0.

Between \$\frac{V_{REF}}{16}\$ and \$\frac{V_{REF}}{16}+\frac{V_{REF}}{8}\$ the output count will be 1 and so on, with each step the same.

When you get up to the top of the range, a count of \$2^N-1\$ (maximum count) will represent a voltage greater than \$V_{REF}\cdot \frac {13}{16}\$.

So- the best guess (minimizing error) for a given count x (where \$ 0 \lt x \lt 2^{N-1}\$) is \$ x \frac{ V_{REF}}{2^{N}}\$, as given in the original question.

Usually MCU ADCs will saturate so ideally in the special case of x=0 you can say the input voltage is less than \$0.5 V_{REF}\cdot \frac{1}{2^N}\$ = \$V_{REF}\cdot \frac{1}{2^{N+1}}\$ and if x = \$2^{N-1}\$ then the input voltage is greater than \$V_{REF}\cdot\frac{ 2^{N+1}-3}{2^{N+1}}\$

Of course with a typical MCU of 10 bits or greater, the ADC error and noise will usually exceed 1 LSB, and Vref will usually not be accurate to 1 LSB either so it doesn't matter that much.

TL;DR

Use the original equation.