For any $n \ge 1$, let $L_n$ denote the minimum number of times the switches must be pressed to turn all the lamps OFF. It is easy to check that $L_1 = 1$ and $L_2 = 2$.

Consider the problem of turning lamps $1, 2, \dots, k$ from all ON to all OFF, for some $k \ge 3$. First, lamp $k$ must be turned OFF at some stage. For this to happen, lamps $1, 2, \dots, k-2$ must all be OFF. This requires a minimum of $L_{k-2}$ switches to be pressed. Then one more switch is required to be pressed in order to turn lamp $k$ OFF. Next consider lamp $(k-1)$. To turn lamp $(k-1)$ OFF, lamp $(k-2)$ must be turned ON. But to turn lamp $(k-2)$ ON, lamp $(k-3)$ must be turned ON, and so on (proof completed by induction). Therefore before lamp $(k-1)$ may be turned OFF, at least as many switches must be pressed as are required to turn all lamps $1, 2, \dots, k-2$ ON - this requires at the minimum a further $L_{k-2}$ switches to be pressed (note that the number of switches required to turn all lamps from OFF to ON is the same as the number required to turn all lamps from ON to OFF). Finally, to turn lamps $1, 2, \dots, (k-1)$ OFF requires a minimum of $L_{k-1}$ switches to be pressed. Summarising,
$$L_k \ge 2L_{k-2} + L_{k-1} + 1, \quad (1)$$
for $k \ge 3$. It is easy to see that the above reasoning defines recursively an algorithm which performs the required switchings while satisfying (1) with equality. Therefore
$$L_k = 2L_{k-2} + L_{k-1} + 1. \qquad (2)$$
Next we prove by induction that for every $r \ge 1$,
$$L_{2r-1} = \frac{4^r - 1}{3} \quad ; \quad L_{2r} = \frac{2(4^r - 1)}{3}$$
This is evident for $r = 1$. Assuming it is true for $r = k$, we have
$$L_{2k+1} = \frac{4^k - 1}{3} \quad ; \quad L_{2k} = \frac{2(4^k - 1)}{3}$$
and so for $r = k + 1$ we have by (2)
$$\begin{aligned} L_{2k+1} &= 2L_{2k-1} + L_{2k} + 1 \\ &= \frac{2(4^k - 1)}{3} + \frac{2(4^k - 1)}{3} + 1 \\ &= \frac{4^{k+1} - 1}{3} \end{aligned}$$
and
$$\begin{aligned} L_{2k+2} &= 2L_{2k} + L_{2k+1} + 1 \\ &= \frac{4(4^k - 1)}{3} + \frac{(4^{k+1} - 1)}{3} + 1 \\ &= \frac{2(4^{k+1} - 1)}{3} \end{aligned}$$

We conclude that for any odd positive integer $n$, we have $n = 2k - 1$, so $k = (n + 1)/2$ and therefore
$$L_n = L_{2k-1} = \frac{4^k - 1}{3} = \frac{4^{\frac{n+1}{2}} - 1}{3} = \frac{2^{n+1} - 1}{3}$$