This can be done very easily without worrying about relativistic addition of velocities because if a charge $q$ crosses a potential difference $V$ then its energy changes by $\Delta E = qV$.
A quick aside: the LHC actually uses RF cavities to accelerate the protons, but let's leave aside that complication and just assume the charged particle repeatedly crosses a potential difference $V$.
The energy of a relativistic particle is:
$$ E^2 = p^2c^2 + m^2c^4 \tag{1} $$
where $p$ is the relativistic momentum:
$$ p = \gamma mv = \frac{mv}{\sqrt{1 - v^2/c^2}} $$
and with a bit of algebra we can rearrange this into an expression for the velocity in terms of the energy:
$$ v = c\,\sqrt{1 - \frac{m^2c^4}{E^2}} \tag{2} $$
So if we start with some energy $E$ and cross a potential difference $V$ the energy changes from $E$ to $E+qV$. So if our initial velocity is:
$$ u = c\,\sqrt{1 - \frac{m^2c^4}{E^2}} $$
then the final velocity is:
$$ v = c\,\sqrt{1 - \frac{m^2c^4}{(E+qV)^2}} $$
And if this occurs over some short distance $\ell$ we can approximate the acceleration using the SUVAT equation:
$$ v^2 = u^2 + 2as $$
to get:
$$ a = \frac{c^2}{2\ell} \left(1 - \frac{m^2c^4}{(E+qV)^2} - 1 + \frac{m^2c^4}{E^2}\right) $$
which for reasons that will shortly become clear I'm going to rearrange to:
$$ a = \frac{m^2c^6}{2\ell} \left( \frac{\frac{qV}{E}\left(2 + \frac{qV}{E}\right)}{E^2\left(1+\frac{qV}{E}\right)^2} \right) \tag{3} $$
You could be forgiven for wondering where on Earth all this algebra is going, but I'm now going to make the assumption that the energy gained in crossing our electric field, $qV$, is much less than the total energy $E$. This is reasonable even for a stationary particle because remember that $E$ includes the rest mass energy so its minimum value is $E+mc^2$ even when the particle is stationary. Since with this assumption $qV/E \ll 1$ our equation (3) simplifies drastically to:
$$ a \approx \frac{m^3c^6}{E^3} \left( \frac{qV}{m\ell} \right) $$
Suppose our particle is stationary then the energy is $E=mc^2$ and our equation becomes:
$$ a_0 = \frac{qV}{m\ell} $$
And this is just the classical acceleration of a charge $q$ in a field gradient $V/\ell$. So we can write our final equation as:
$$ a = \frac{m^3c^6}{E^3} a_0 \tag{4} $$
where $a_0$ is the acceleration at non-relativistic speeds. And there's our result. The acceleration we observe in the lab is less than the acceleration at non-relativistic speeds by a factor of $m^3c^6/E^3$.
If you haven't completely lost the will to live by now there is one more step we can take. For very highly relativistic particles the energy is much greater than the rest mass and our original energy equation (1) becomes:
$$ E\approx pc \approx \gamma mv$$
Substituting this in our equation (4) we get:
$$ \frac{a}{a_0} \approx \frac{c^3}{\gamma^3v^3} $$
And since for highly relativistic particles $v \approx c$ we end up with:
$$ \frac{a}{a_0} \approx \frac{1}{\gamma^3} \tag{5} $$
