it would be convenient if there was a way to synchronize the clocks perfectly (or near as darn it)
The easier way is to equip the lattice clocks with radio transmitters and receivers and synchronize using light pulses.
However, slow clock transport is interesting in its own right.
does the total amount that the master clock get set back go down rapidly as the velocities and accelerations go down, even though the journey time goes up correspondingly?
To travel a distance $L$ at a constant speed $v$ requires a time $t=L/v$. The accelerations to turn do not affect time dilation provided the speed is constant. During time $t$ the clock loses a time $$\Delta t=(\gamma-1)t$$ where $\gamma$ is the time dilation factor as defined in the question.
Now, if we are interested in the behavior at very small $v$ then we can do a first-order Taylor series expansion about $v=0$. Recall the first order expansion about $v=0$ is $$ f(v)= f(0) + f'(0)\ v + O(v^2)$$ where $O(v^2)$ is all of the terms of order $v^2$ or higher. Those terms become negligible as $v$ becomes small. Using Mathematica to evaluate the Taylor series gives $$\Delta t=\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)\frac{L}{v}=\frac{Lv}{2c^2}+O\left(v^2\right)$$ so for small $v$ the time error is proportional to $L$ and $v$. For a fixed $L$ it can therefore be made arbitrarily small by using a small $v$.
This could be reasonable for a small lattice, but if the lattice is a 3D lattice then $L$ would grow as the cube of the lattice size, $s$ (assuming lattice spacing remains constant). Since $L\propto s^3$, to keep a fixed $\Delta t$ would require $v = 2c^2 \ \Delta t/L \propto s^{-3}$ to decrease by the inverse cube of the lattice size. So $t=L/v\propto s^3/s^{-3}=s^6$ would grow as the 6th power of the lattice size! That would rapidly become infeasible even for modest size lattices. The radio approach would become rapidly more appealing.