From your setup, there is symmetry to exploit. If you break up the T-to-3T trip in half, you can see [by drawing a spacetime diagram] that you have four congruent legs to your trip. You just calculated the duration of the first leg. So, by symmetry, multiply that by 4 to get the total duration.
In this second update, I have shown three uniformly accelerated observers as with dashed worldlines, which are hyperbolas on a spacetime diagram.
Since the magnitudes of their proper-accelerations are equal, these hyperbolas are congruent--related by reflections in space, reflections in time, and translations on this spacetime diagram.
[If we were doing Galilean relativity, these would be congruent parabolas.]
This round-trip starting and ending at rest in this frame is essentially
the splicing of four congruent portions.
[As a simpler case, suppose the traveler departed with a nonzero outgoing speed
and returned at the same speed.... essentially the magenta observer. Is it clear that if that trip were split into halves, those halves have equal elapsed times.]
Note that the 0-to-T leg is the mirror image in space of the 2T-to-3T leg.
Note that the T-to-2T leg is the mirror image in space of the 3T-to-4T leg.
The 0-to-T leg is the "time-reverse" of the 3T-to-4T leg.
The T-to-2T leg is the "time-reverse" of the 2T-to-3T leg.
Consider the first leg of the trip... uniform-acceleration $a_0$ to the right.
Suppose instead the traveler went in the opposite direction in space.
After the same time $\Delta t_1$, is it clear that the traveler would have the opposite velocity (same speed, opposite direction) as the original?
This space-reflected segment is a translation of the leg that starts [back] at the turnaround event.
Similarly, if you ran time in reverse [retrodiction, if you will], then at earlier time $-\Delta t_1$,
is it clear that the traveler would have had the opposite velocity (same speed, opposite direction) as the original?
This time-reflected segment is a translation of the leg that returns at the reunion event.