Lower bound on algorithm solving certain recurrence
There's quite a lot of confusion here:

$\Omega(n)$ and $\Omega(n-1)$ are absolutely the same.
You don't an adversary argument to show that this algorithm runs in $\Theta(n)$. It follows immediately from the description of the argument.
There is a difference between analyzing the running time of a particular algorithm and showing that every algorithm that correctly solves a problem must run in at least so many steps.

In order to show that any algorithm computing $F_n$ correctly must run in time $\Omega(n)$, we will show that any such algorithm must read all inputs $p_1,\ldots,p_n$. The starting point is the following observation:

The correct output is different for the instances $p_1=\cdots=p_n=0$ and $p_1=\cdots=p_{i-1}=0,p_i=1,p_{i+1}=\cdots=p_n=0$.

Given that, we can use a simple adversary argument to show that any algorithm correctly computing the function must read all $p_j$. Indeed, run the given algorithm, answering $0$ whenever it queries the value of some $p_j$. If it announces the answer without querying, say, $p_i$, then it cannot tell apart the two instances above, and in particular, its answer will be wrong on at least one of them.