Answer
$ f=\dfrac{1}{2\pi}\sqrt{\dfrac{2T}{mL}} $
Work Step by Step
Since the amplitude is sufficiently small and the magnitude of the tension in the rubber bands is essentially unchanged, then we are dealing with a small angle of oscillation. See the figure below.
Now we need to apply newton's second law on the block in the $y$-direction. (Noting that the $x-y$ plane is a horizontal plane).
$$\sum F_y=T\sin\theta+T\sin\theta=ma_y$$
$$2T\sin\theta=ma_y\tag 1$$
We can see that the net $x$-component of the force is zero since they have the same magnitude and opposite directions.
From the right triangle, which is the second figure below, we can see that $$\sin\theta=\dfrac{y}{r}$$ \and $r=\sqrt{L^2+y^2}$, so
$$\sin\theta=\dfrac{y}{\sqrt{L^2+y^2}}$$
And since we are dealing with a very small angle, then $y\lt\lt L$ which means that $y^2$ approaches zero, so $\sqrt{L^2+y^2}\approx \sqrt{L^2}=L$.
Hence,
$$\sin\theta=\dfrac{y}{L}$$
which is also $\tan\theta$ which means that our angle $\theta\lt\lt 1\;\rm rad$
Plugging that into (1);
$$2T\dfrac{y}{L}=ma_y $$
Recall that $a_y=d^2y/dt^2$, so
$$\dfrac{2T}{L} \;y=m \dfrac{dy^2}{dt^2} $$
$$\dfrac{dy^2}{dt^2}=\dfrac{2T}{mL} \;y $$
we can see that this formula is equivalent to $dx^2/dt^2=-kx/m$.
Thus, $\dfrac{-k}{m}=\dfrac{2T}{mL}=\omega^2$ where $\omega=2\pi f$.
Hence,
$$\omega=\sqrt{\dfrac{2T}{mL}}=2\pi f$$
Therefore,
$$\boxed{f=\dfrac{1}{2\pi}\sqrt{\dfrac{2T}{mL}}}$$
where $T$ here is the tension force, not the period.
