The result is based on a recurrence.

First I am showing you the example for understanding and then prove it.

Suppose, you have to design $3$-input NOR gate. So, $n=3.$ Suppose, inputs are $A,B,C$ and So, result would be:

$\overline{A+B+C} = \overline{\overline{\overline{A+B}}+C}$

Here, $\overline{A+B}$ requires one 2-input NOR gate, Now, $\overline{\overline{A+B}}$ requires total two 2-input NOR gates (1 NOR gate from previous $\overline{A+B}$) and finally, $\overline{\overline{\overline{A+B}}+C}$ requires total threee 2-input NOR gates (2 NOR gates from previous $\overline{\overline{A+B}}$).

So, you can verify $3$ is actually $2*3-3.$

Now, Based on this, you can prove the result as:

Suppose,

$ \overline{A_1+A_2+A_3+…+A_n} = \overline{\overline{\overline{A_1+A_2+A_3+...+A_{n-1}}}+A_n} $

Let, $T_n$ be the minimum number of two-inputs NOR gates. Then, you can write the recurrence as:

$T_n = T_{n-1} + 1 + 1$

i.e. $T_n = T_{n-1} + 2$ with $T_2 = 1$

Hence, $T_n = 2n-3$