There are lots of different examples of oscillatory systems that have essentially the same mathematical form. Let's start by just looking at one type of differential equation:
$a = \frac{d^2 x}{dt^2} = -\omega^2 x$
This equation has a general solution (you can check this)
$x(t) = A \sin (\omega t + \phi)$
which oscillates with a period of $T=2\pi/\omega$ since the system will be in exactly the same state at any time $t$ and $t + 2\pi/\omega$. So now if we find physical systems that are described by an acceleration equation that looks like this, we know exactly what the solution is.
For example, in a mass spring system we would have
$ma=-kx \rightarrow \frac{d^2 x}{dt^2} = -\left(\frac{k}{m}\right) x$
which is exactly the same form as we had before with $\sqrt{\frac{k}{m}}=\omega$, so plugging this into our equation for period we get $T=2\pi\sqrt{\frac{m}{k}} $
The same applies for your torsion pendulum, you switch the position $x$ with angular position $\theta$ (this doesn't change how to solve the differential equation) and have
$\frac{d^2\theta}{dt^2} = -\left(\frac{C}{I}\right) \theta = -\omega^2 \theta$
as your differential equation with $\omega = \sqrt{\frac{C}{I}}$, it has the same solution, and plugging this $\omega$ into your period formula you get
$T=2\pi \sqrt{\frac{I}{C}}$
This shows that once you solve the general form of one type of differential equation you can apply it to any other type of system that has the same form for its equation (this is also used in circuits, perturbations to an orbit, or pretty much anything else that is slightly pushed away from a stable equilibrium position)
Since your pendulum oscillates about a new angle, call it $\theta_0$, your Taylor series approximation should be about $\theta_0$. So,
\begin{align*}
l \ddot{\theta} & = -g \sin{\theta} + A \cos{\theta} \\
& \approx -g ( \sin{\theta_0} + (\theta - \theta_0) \cos{\theta_0} ) + A ( \cos{\theta_0} - (\theta - \theta_0) \sin{\theta_0}) \text{ for small deviations from } \theta_0 \\
& = A \cos{\theta_0} - g \sin{\theta_0} + \theta_0 (g \cos{\theta_0} + A \sin{\theta_0}) - \theta (g \cos{\theta_0} + A \sin{\theta_0})
\end{align*}
Solving that should give you the period you're looking for.
Best Answer
Gravity exerts a constant force downwards. If you apply a second constant force sideways you are in effect rotating the gravitational force:
where the angle $\theta$ is given by:
$$ \tan\theta = \frac{F_\text{ext}}{F_g} $$
The modulus of the net force is given by the usual Pythagoras rule:
$$ F_\text{net}^2 = F_g^2 + F_\text{ext}^2 $$
So if you imagine mentally rotating the whole experiment an angle $\theta$ anticlockwise you would have the pendulum handing vertically downwards in an effective gravitational field of $F_\text{net}$. This should make it obvious how the period changes.