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I am not sure if this answers your question but I hope it could be meaningful and leads to some insight.

Assume that there is a turing machine $X$ that can simulate every atom in the universe including itself, it then necessarily can simulate itself.

Now, reducing that to the halting problem is trivial:

Let $X$ take a turing machine $M$ as its input and decides whether it halts or not by simulating the universe (since $M$ is included in the universe), then do the opposite (e.g. $X$ halts if $M$ does not, and loops forever if $M$ halts). Then $X(X)$ demonstrates a contradiction.

Essentially this means that the best $X$ can do to decide whether $X$ halts or not is just by running itself (i.e. let the universe work its way), so simulating the universe doesn't give an advantage.

The same applies when you want the state of the universe after $t$ time. Since $X$ can not decide if it will halt within $t$ time or not within $t$ time (same argument), then it will let it to the universe to do it. Trying to simulate the universe doing it, can not reduce the time you'll take to decide. And if deciding how the universe will look like in $t$ time takes more than $t$ then the simulation will diverge (as $t$ goes to infinity).

This leads to the conclusion that only useful simulator that decides how the universe will look like in $t$ time must take exactly $t$ time, i.e. by letting the universe work. This simulator is then indeed the trivial simulator.