Number of Final States Subset Construction for NFA to DFA
Think about which states in the new NFA don't contain a state that was in $F$. These are   $\emptyset$ and every combination of states from $\mathcal{Q}-F$, i.e. $\mathcal{P}(\mathcal{Q}-F)$.

In your 3-state example, the only elements of $\mathcal{P}(\{1,2,3\})$ that don't contain 1 or 2 are $\emptyset$ and $\{3\}$, which constitute $\mathcal{P}(\{3\})$, with cardinality 2, so we have $\mid F'\mid=8-2=6$.

For your 7-state example with $\mid F\mid=2$, the size of $\mathcal{P}(\mathcal{Q}-F)$ is 32, which means there are $\mid F'\mid=128-32=96$ states that also contain at least one state that was in $F$.

So your conjecture of $2^{\mid F\mid}-2^{\mid\mathcal{Q}-F\mid}$ is correct.