Theoretical precision needed to get $n$-bits of the evaluation of some sum
I may be misunderstanding the question, but I don't think we have enough information to tell.

Consider, for example:

$$A = \left\{ \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \right\}$$

and:

$$P(x) = \left( x - \frac{1}{4} - \varepsilon \right) \left( x - \frac{1}{2} + \varepsilon \right) \left( x - \frac{3}{4} - \varepsilon \right)$$
$$Q(x) = 1$$

for some small $\varepsilon$.

Assuming that $\varepsilon$ is small enough, both polynomials are height-bounded by 1.

The sum is:

$$S = - \frac{5}{16}\varepsilon + \varepsilon^3$$

The question is how you would evaluate $P$.

$$P(x) = \left(-\frac{3}{32} - \frac{5}{16}\varepsilon + \frac{1}{2}\varepsilon^2 + \varepsilon^3\right) + \left( \frac{11}{16} + \varepsilon - \varepsilon^2\right) x + \left(-\frac{3}{2} - \varepsilon\right) x^2 + x^3$$

The number of bits of precision on the elements of $A$ are not relevant here; they are precisely representable with two significant bits. It seems to me that the precision you need to do the intermediate computation is more important.

For a small enough $\varepsilon$, you are going to get catastrophic cancellation whether you use the factorised form of $P$ or you use a method such as Horner's rule. At the very least, you need enough bits to represent $-\frac{3}{32} - \frac{5}{16}\varepsilon$ such that you have $n$ significant bits in the $\varepsilon$ term, so you can subtract $-\frac{3}{32}$ to get a good enough answer.