The answer is $16n - 8$.
Assume that $P_i = (x_i, y_i)$ for $1 \le i \le 4n + 1$. Set
$$P_{4n+2} = (x_{4n+2}, y_{4n+2}) = P_1 = (x_1, y_1).$$
We will show that the sum
$$S = \sum_{i=1}^{4n+1} (P_i P_{i+1})^2 \geq 16n - 8.$$
First, we show that this minimum can be obtained by setting
$$(P_1, P_2, \dots, P_{4n+1}) = ((2, 0), (4, 0), \dots, (n, 0), (n-1, 0), \dots, (1, 0), (0, 2), (0, 4), \dots, (0, n), (0, n-1), \dots, (0, 1), (-2, 0), (-4, 0), \dots, (-n, 0), (-n+1, 0), \dots, (-1, 0), (0, -2), (0, -4), \dots, (0, -n), (0, -n+1), \dots, (0, -1), (0, 0))$$
for $n$ even, and
$$(P_1, P_2, \dots, P_{4n+1}) = ((1, 0), (3, 0), \dots, (n, 0), (n-1, 0), \dots, (2, 0), (0, 1), (0, 3), \dots, (0, n), (0, n-1), \dots, (0, 2), (-1, 0), (-3, 0), \dots, (-n, 0), (-n+1, 0), \dots, (-2, 0), (0, -1), (0, -3), \dots, (0, -n), (0, -n+1), \dots, (0, -2), (0, 0))$$
for $n$ odd. This can be easily checked. For example, when $n = 2m$ is even, our construction shows that
$$\begin{aligned} S &= 4 \left( \sum_{i=1}^{m-1} (P_i P_{i+1})^2 + (P_m P_{m+1})^2 + \sum_{i=m+1}^{2m-1} (P_i P_{i+1})^2 \right) \\ &\quad + 3(P_{2m} P_{2m+1})^2 + (P_{4n} P_{4n+1})^2 + (P_{4n+1} P_1)^2 \\ &= 4(4(m-1)+1+4(m-1)) + 3 \times 5 + 1 + 4 \\ &= 32m - 8 = 16n - 8. \end{aligned}$$
The exact same argument works when $n$ is odd.
Note that
$$S = \sum_{i=1}^{4n+1} [(x_i - x_{i+1})^2 + (y_i - y_{i+1})^2].$$