There are no such solutions!
To see this explicitly we can calculate how many cookies the 3 people ( A, B and C ) have at a given stage. To do this I altered the problem by assuming that after person A (Alice) shares their cookies for the first time, everyone gives their cookies to the person alphabetically before them: so C gives to B, B gives to A and A gives to C. In practice this gives the following for the first 2 stages:
At stage 1: A has n, B has 0 and C has 0
At stage 2: A has n/2, B has n/2 and C has 0
We can repeat this procedure of A halving the cookies and then the cookies being shifted along (so now only Alice halves her cookies). This generates the same fractions to appear at each stage as the original problem yet different people will have different fractions. (The shift essentially changes A into person B on stages that are 2 mod 3, C on 0 mod 3 stages and A on 1 mod 3 stages)
(note C will always have 0 cookies)
From this we can calculate how many cookies people have at a given stage in the new altered problem (and this will be a permutation of the solution to the original problem as discussed above).
Let A(i),B(i) denote the cookies A and B have a stage i.
We have $A(1) = n$ and $B(1) = 0 $
We now consider how A(i) changes in the altered problem, we deduce:
$A(i+1) = B(i) + \frac {A(i)} {2} $
and $B(i+1) = \frac{A(i)} {2} $
Therefore $B(i) = \frac{A(i-1)} {2}$ from that second equation
Substituting this back in to the first gives (after arranging):
$2A(i+2) - A(i+1) - A(i) = 0$
This is a standard difference equation which we can solve with the initial conditions I wrote above. I'll leave out the algebra. This gives:
$ A(i) = \frac {2n}{3} \frac{2^i - (-1)^i} {2^i}$
and since no cookies are eaten B(i) = n - A(i)
And from this we know how many cookies A has in the altered problem which implies someone in the original problem has A(i) cookie at a given stage. We now note that A(i) must always be an integer for the cookies to be halved correctly without one being eaten.
However it is clear that $(2^i - (-1)^i)$ is always an odd number so for A(i) to be an integer for all i, 2^i must divide n for all positive intgers i. This is not possible unless we are in the trival case in which n=0.
(note it is possible to calculate exactly how many cookies each person has at a stage by combining the solution to the difference equation with my comment :'(The shift essentially changes A into person B on stages that are 2 mod 3, C on 0 mod 3 stages and A on 1 mod 3 stages)')