The simple answer: for anything further than the moon, you should use FTL.
The slightly more technical answer: 18 million km is the break-even point.
Obviously, price is going to have a large affect on what you actually use FTL for, but if speed is your worry, FTL is already 3 times faster by Mars (at closest approach), and just gets better from there. By the time you're sending stuff to Proxima, there's a 12300 times speed increase (it doesn't quite hit 12500 because of the acceleration times).
Short Derivation
The closest planet-like object beyond the moon is Mars. At its closest approach, it's 56 million km from Earth. That's 3 minutes, 7 seconds by light.
Our hard drives can be processed in less than a minute, hit max speed in less than a minute, then decelerate in less than a minute. That's less than 3 minutes. Ignoring the fact that the drive is moving during acceleration (reducing the time even more), at 12.5k c, the 56 million km takes 15 ms.
Less than 3 minutes + 15 milliseconds is less than the 3 minutes and 7 seconds it takes light at closest approach. For any other part of the orbit, the difference is even greater. For any other planet or moon, the difference isn't even close.
Longer Derivation
I'll use exactly 60 seconds for processing time since "less than" is annoying. Then, let's assume constant acceleration to/from 12500 c.
$A(t)=a$
$V(t)=\int{A(t)}dt$$=at+v_0$$=at$
$D(t)=\int{V(t)}dt=\frac{a}{2}t^2+v_0t+d_0$$=\frac{a}{2}t^2$
Where $a$ is constant acceleration, $v_0$ is initial velocity (zero here), $d_0$ is initial position (which we can call zero), and A, V, D are functions of acceleration, velocity, and distance.
First, let's solve for $a$:
$V(t)=12500c$$=3.747\cdot10^{12} \frac{m}{s}$$=a\cdot60s$
$a=\frac{1}{60s}\cdot3.747\cdot10^{12}\frac{m}{s}=$$6.245\cdot10^{10}\frac{m}{s^2}$
(About 6 billion gees.)
Now, how far does the drive go before hitting max velocity?
$d=\frac{6.245\cdot10^{10}}{2}\frac{m}{s^2}(60s)^2=$$1.124\cdot10^{14}m$
Which is about 8 times the distance to the edge of the solar system. Since your furthest colony, Proxima b, is 4.2 ly ($3.978\cdot10^{16}m$) away, we need to use piecewise functions for it, but for everything else, we'll never see max velocity.
We can solve for the time taken to get to any specific planet. Since (I presume) we have to accelerate then decelerate, take the half-distance time and multiply by 2.
$\frac{d}{2}=\frac{a}{2}t_\text{half}^2$$\rightarrow t_\text{half}=\sqrt{\frac{d}{a}}$$\rightarrow t=2t_\text{half}=2\sqrt{\frac{d}{a}}$
Specific Travel Times
$t_\text{MarsClose}=2\sqrt{\frac{56\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$1.894 s$
(plus 60 seconds for processing is 62ish seconds compared to 182 seconds at light speed)
$t_\text{MarsAverage}=2\sqrt{\frac{225\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$3.796 s$
(64 seconds compared to 751 seconds)
$t_\text{MarsFar}=2\sqrt{\frac{401\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$5.068 s$
(65 seconds compared to 1342 seconds)
$t_\text{JupiterNear}=2\sqrt{\frac{588\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$6.137 s$
(66 seconds compared to 1961 seconds)
$t_\text{JupiterFar}=2\sqrt{\frac{968\cdot10^{9}m}{6.245\cdot10^{10}\frac{m}{s^2}}}=$$7.874 s$
(68 seconds compared to 3229 seconds)
$t_\text{Proxima}=2\sqrt{\frac{1.124\cdot10^{14}m}{6.245\cdot10^{10}\frac{m}{s^2}}}+\frac{3.978\cdot10^{16}m-1.124\cdot10^{14}m}{3.747\cdot10^{12}\frac{m}{s}}=$$10671 s$
(about 3 hours compared to about 4.2 years)
Exact Point of Equality
We can calculate the exact distance where the two times are equal. $p$ is the 60 seconds of processing time.
$\frac{d}{c}=2\sqrt{\frac{d}{a}}+p$
(Mumble mumble stupid sign errors doing it by hand, just plug it into WolframAlpha /mumble mumble.)
$d=\frac{1}{a}[acp+2c^2\pm2\sqrt{ac^3p+c^4}]$$=cp+\frac{2c^2}{a}\pm\frac{2}{a}\sqrt{ac^3p+c^4}$
$cp+\frac{2c^2}{a}$$=3\cdot10^8\frac{m}{s}\cdot60s+\frac{2(3\cdot10^8\frac{m}{s})^2}{6.245\cdot10^{10}\frac{m}{s^2}}=$$1.8\cdot10^{10}m$
$\frac{2}{a}\sqrt{ac^3p+c^4}$$=\frac{2}{6.245\cdot10^{10}\frac{m}{s^2}}\sqrt{6.245\cdot10^{10}\frac{m}{s^2}\cdot60s\cdot(3\cdot10^8\frac{m}{s})^3+(3\cdot10^8\frac{m}{s})^4}=$$3.221\cdot10^8m$
Since the second term is tiny compared to the first term, we can pretty much ignore it, but the exact distances where the two meet are:
$d_+=$$1.832\cdot10^{10}m$
$d_-=$$1.768\cdot10^{10}m$
Plotting the curves on a graph, I only see an intersection at $d_+$, so I'd go with that one.
$1.8\cdot10^{10}m$ is $18$ $\text{million}$ $km$, or $1.019$ light minutes, which makes sense, given the 60-second processing time and how ridiculously fast our FTL accelerates.
Conclusion
For anything further than 18 million km, FTL is faster. That's a lot further than the moon (about 0.39 million km), but nowhere near Mars (56 million km at closest approach) or Jupiter (588 million km at closest approach).
