How does the order of Q affect the time it takes to solve ECDLP?
Here is what I believe Sagemath is doing:


Given the problem of finding $x$ s.t. $xP = Q$, one step is computing $x' = x \bmod 5888291787538299579505114081452341011$; it does this by computing $P' = (2\cdot 3 \cdot 5)P$, and $Q' = (2 \cdot 3 \cdot 5)Q$, and trying to solve $x'P' = Q'$
As the order of $P'$ is circa $2^{122}$, we would normally expect this computation to take circa $2^{61}$ time (quite a while).  However, in this case, $Q'$ is the neutral element.  I suspect that Sagemath notices this, and so immediately concludes that $x' = 0$
Every other computation involved in Pohlig-Hellman is relatively quick, and so it reports the answer ($3 \cdot 5 \cdot 5888291787538299579505114081452341011$) quickly.


The summary is: if your implementation special cases this $Q' = 0$ case, then we can skip any prime that does not appear in the order of $Q$; the time taken by Polhig-Hellman is dominated by the largest prime that appears both in the order of $P$ and the order of $Q$.