Finding the fundamental frequency of a periodic signal
1) You are almost there if you have found an $ l,m,n$ and a $T_0$ that fit your equations.  Let $p$ be the largest common factor of $ l,m,n$.  If $p=1$ you are done, otherwise divide $ l,m,n$ and $T_0$ by $p$ and now you are done.

2) From your own equations:

$$ \omega_1 = \omega_0 \cdot l $$
$$ \omega_2 = \omega_0 \cdot m $$
$$ \omega_3 = \omega_0 \cdot n $$

You already know that $l,m,n$ don't have a common factor greater than one.

Hope this helps.

Ced

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Followup:

Taking it from the top.  You are supposing that you know the frequency ratios:

$$ \frac{\omega_1}{\omega_2} = \frac{a}{b} $$
$$ \frac{\omega_2}{\omega_3} = \frac{c}{d} $$

So $a,b,c,d$ are known.  Therefore:

$$ \frac{l}{m} = \frac{a}{b} $$
$$ \frac{m}{n} = \frac{c}{d} $$

Cross multiply:

$$ lb = ma $$
$$ md = nc $$

Let $ m = bc $ and therefore $ l = ac $ and $ n = bd $

You now have an initial $l,m,n$ set that you can possibly reduce as stated above.  Then the fundamental frequency, $ \omega_0 $, can be solved for using any of the three values.  

$$ \omega_0\ = \omega_1 / l = \omega_2 / m = \omega_3 / n $$

Finally, $T_0 = 2\pi / \omega_0 $ gives you the period.

Quite Easily Done

Note, it is possible that none of $l,m,n$ are equal to one.