Cube the digits and carry on

Considering cycles

The largest number such a chain can ever reach is $1486$ (every number between $2001$ and $2100$ gives at most $8+0+729+729=1466$ at the first step, and the largest possibility resulting from any number up to there is $1+27+729+729=1486$). So we have an upper bound, which means every chain must eventually end in a cycle.

In the OP you mentioned $overline{153}$ as a cycle of length 1; there are only five cycles of length 1, namely $overline{1}$, $overline{153}$, $overline{370}$, $overline{371}$, $overline{407}$. I found most of these and also a length-three cycle $overline{133rightarrow55rightarrow250}$ just by experimenting with starting with single-digit numbers:

• $9rightarrow729rightarrow1080rightarrow513rightarrowoverline{153}$

• $8rightarrow512rightarrow134rightarrow92rightarrow737rightarrow713rightarrowoverline{371}$

• $7rightarrow343rightarrow118rightarrow514rightarrow190rightarrow730rightarrowoverline{370}$

• $6rightarrow216rightarrow225rightarrow141$, goes to $overline{153}$ as in the OP.

• $5rightarrow125$, goes to $overline{371}$ as for $8rightarrow512$ above.

• $4rightarrow64rightarrow280rightarrow520rightarrowoverline{133rightarrow55rightarrow250}$

• $3rightarrow27rightarrow351rightarrowoverline{153}$

• $2rightarrow8$, goes to $overline{371}$ as seen above.

• $overline{1}$ is of course its own fixed cycle.

Eliminating cases

The chain you've given in the OP also gives a bunch of numbers that must end with $overline{153}$:

Knowing that $9$ goes there also gives:

However, from $8$ and $4$ we also find that

So far, among the 100 given starting points, we know that 13 of them do go to $overline{153}$ and 6 of them don't. We also know the ending point must be a cycle, either one of the five possible single-point cycles or a multi-point cycle; at least one multi-point cycle is possible, but I don't know how many are.

As Jaap Scherpuis astutely notes, the only possible numbers which could go to $overline{153}$ are

We already know 13 of these do go to $overline{153}$, so that leaves 21 left to check. Checking the first couple of them:

Now we already get ten more for free:

Next one:

giving us four more for free:

Only seven left to check, namely

We just need to check four of these:

Final solution