I agree with you.

It is easier to amend the game slightly, so that in a round all three players reveal their pronouncement at the same time, independent of one another. This simplifies the reasoning because it is no longer turn based, so the order of the players no longer matters.

I think that this version is equivalent in the sense that in the end the same conclusions will be reached by the players. If you compare a single round of this version with one round around the table of the original, the players have less information to base their deductions on, so this version is at least as hard for the players as the original. On the other hand, comparing a single player's turn of the original to one round of the new version shows that more information is revealed in the new version, so the original is at least as hard as the new one. I'm not totally convinced that reasoning is valid, but for now lets assume it is.

In this simultaneous-play version of the game, deductions will be as follows:

Round 1: If all players have a pair (AA or 88), then one player will see four the same cards and know their hand. If that happens, the other two players will be able to deduce their hand and announce it in the next round.
If no one deduces their hand, this round eliminates the no-mixed-hand cases AA/AA/88 and AA/88/88.

Round 2: If any player sees two pairs (AA and 88), then they will know they have a mixed pair (A8). If that happens, the other two players will be able to deduce their hand and announce it in the next round.
If no one deduces their hand, this round eliminates the one-mixed-hand case AA/88/A8.

Round 3: If a player sees one pair (AA/A8 or 88/A8), then they will know they have a mixed hand (A8). If that happens it will be to two players, but the player with the matched pair will not know whether he has AA or 88.
If no one deduces their hand, this round eliminates the two-mixed-hand cases AA/A8/A8 and 88/A8/A8.

Round 4: Now everyone knows it must be the three-mixed-hand case A8/A8/A8, and all players deduce their hand.

So at worst one player cannot deduce their hand because they see A8/A8 and can have either AA or 88.

[EDIT: The simplification I apply below fundamentally alters the game, so cannot really tell us anything about the original game. See BartBog's answer for why.]

I agree with you.

It is easier to amend the game slightly, so that in a round all three players reveal their pronouncement at the same time, independent of one another. This simplifies the reasoning because it is no longer turn based, so the order of the players no longer matters.

I think that this version is equivalent in the sense that in the end the same conclusions will be reached by the players. If you compare a single round of this version with one round around the table of the original, the players have less information to base their deductions on, so this version is at least as hard for the players as the original. On the other hand, comparing a single player's turn of the original to one round of the new version shows that more information is revealed in the new version, so the original is at least as hard as the new one. I'm not totally convinced that reasoning is valid (Edit:- It isn't!) , but for now lets assume it is.

In this simultaneous-play version of the game, deductions will be as follows:

Round 1: If all players have a pair (AA or 88), then one player will see four the same cards and know their hand. If that happens, the other two players will be able to deduce their hand and announce it in the next round.
If no one deduces their hand, this round eliminates the no-mixed-hand cases AA/AA/88 and AA/88/88.

Round 2: If any player sees two pairs (AA and 88), then they will know they have a mixed pair (A8). If that happens, the other two players will be able to deduce their hand and announce it in the next round.
If no one deduces their hand, this round eliminates the one-mixed-hand case AA/88/A8.

Round 3: If a player sees one pair (AA/A8 or 88/A8), then they will know they have a mixed hand (A8). If that happens it will be to two players, but the player with the matched pair will not know whether he has AA or 88.
If no one deduces their hand, this round eliminates the two-mixed-hand cases AA/A8/A8 and 88/A8/A8.

Round 4: Now everyone knows it must be the three-mixed-hand case A8/A8/A8, and all players deduce their hand.

So at worst one player cannot deduce their hand because they see A8/A8 and can have either AA or 88.