Powering a buck/boost converter, how much amps can I get?
If we have to think in watts, would it mean that...
  
  PSU: 12 * 3.2 = 38.4 Watts
  
  LTC3780 @ 5V: 38.4 / 5 = 7.68 A


Assuming 100% efficiency, yes. 

Your calculations are correct and makes sense. 



\$P_{efficiency}=\frac{P_{out}}{P_{in}}\$, you are saying that this ratio is equal to 1.

If we plug in some numbers... we get this: 

$$
\begin{align}
1 &amp;= \frac{P_{out}}{38.4W}\\\\
38.4W &amp;= P_{out}\\\\
38.4W &amp;= V_{out}×I_{out}\\\\
I_{out}&amp;=\frac{38.4W}{ V_{out}}
\end{align}
$$

So this will be your maximum current draw at any given voltage. 

For P to be the same and impedance is constant, then with lower voltage the current increases, or higher voltage and lower current. 

But... in the real world there's no such thing as 100% efficiency... Also,  what you will notice is that it will be hard for you to get like 1000 A and 0.0384 V. But it will be relatively easier to get 1000 V and 0.0384 A. This is because there's always parasitic resistances everywhere. Except for in superconducting materials (as far as I know). 



There's not much else to say really.