If we represent T with -1 and F with 1, the polynomial corresponding to the Boolean AND function $x_1 \land x_2$ is:
$p(x_1, x_2) = \dfrac{1}{2} \left( x_1 + x_2 - x_1 x_2 + 1 \right)$
We can verify by checking that $p(-1, -1) = -1$, while $p(-1, 1) = p(1, -1) = p(1, 1) = 1$.
Similarly, the polynomial for the AND function over three variables $x_1 \land x_2 \land x_3$ is:
$p(x_1, x_2, x_3) = \dfrac{1}{4} \left( x_1 + x_2 + x_3 - x_1 x_2 - x_1 x_3 - x_2 x_3 + x_1 x_2 x_3 + 3 \right)$
And for four variables $x_1 \land x_2 \land x_3 \land x_4$, we have:
$p(x_1, x_2, x_3, x_3) = \dfrac{1}{8} \left( x_1 + x_2 + x_3 + x_4 - x_1 x_2 - x_1 x_3 - x_1 x_4 - x_2 x_3 - x_2 x_4 - x_3 x_4 + x_2 x_3 x_4 + x_1 x_3 x_4 + x_1 x_2 x_4 + x_1 x_2 x_3 - x_1 x_2 x_3 x_4 + 7 \right)$
How can we express this pattern as a closed form? I have the basic framework
$p(x_1, \cdots, x_n) = \dfrac{1}{2^{n-1}} \left( \sum_{i=1}^n x_i + \cdots + 2^{n-1} - 1 \right)$
but am unsure what to put in the middle in order to capture the changes of sign and the increasing number of variables.
We can use the fact that, if we represent the booleans as elements of $\{0,1\}$ instead of $\{-1,1\}$ (with TRUE=$0$ and FALSE=$1$), and is equivalent to simple multiplication. Using the map $f(x)=\frac{1-x}{2}$ sending $-1\to 1$ and $1\to 0$, we may write \begin{align}\text{AND}(x_1,x_2,\dots,x_n) &=f^{-1}\big(f(x_1)f(x_2)\cdots f(x_n)\big)\\ &=f^{-1}\left(\prod_{i=1}^n \frac{1-x_i}{2}\right)\\ &=1-2\prod_{i=1}^n \frac{1-x_i}{2}. \end{align} You can write this in many different forms.