First we prove that, for any positive real numbers $p_1, \dots, p_n, q_1, \dots, q_n, r_1, \dots, r_n$, the following inequality holds:
$$(p_1^3 + \cdots + p_n^3)(q_1^3 + \cdots + q_n^3)(r_1^3 + \cdots + r_n^3) \ge (p_1q_1r_1 + \cdots + p_nq_nr_n)^3.$$
Indeed, by using Cauchy-Schwarz inequality repeatedly, we obtain
$$\begin{align*} & (p_1^3 + \cdots + p_n^3)(q_1^3 + \cdots + q_n^3)(r_1^3 + \cdots + r_n^3)(p_1 q_1 r_1 + \cdots + p_n q_n r_n) \\ \ge & ((p_1^{\frac{3}{2}} q_1^{\frac{3}{2}} + \cdots + p_n^{\frac{3}{2}} q_n^{\frac{3}{2}})(p_1^{\frac{1}{2}} q_1^{\frac{1}{2}} r_1^2 + \cdots + p_n^{\frac{1}{2}} q_n^{\frac{1}{2}} r_n^2))^2 \\ \ge & ((p_1 q_1 r_1 + \cdots + p_n q_n r_n)^2)^2 \\ = & (p_1 q_1 r_1 + \cdots + p_n q_n r_n)^4 \end{align*}$$
and hence the desired inequality.

Let $t = 6^{-\frac{1}{3}}$. Using this inequality and AM-GM inequality,
$$\begin{align*} & (x_1^3 + x_2^3 + x_3^3 + 1)(y_1^3 + y_2^3 + y_3^3 + 1)(z_1^3 + z_2^3 + z_3^3 + 1) \\ &= (x_1^3 + x_2^3 + x_3^3 + t^3 + t^3 + t^3 + t^3 + t^3 + t^3) \\ & \quad \times (t^3 + t^3 + t^3 + y_1^3 + y_2^3 + y_3^3 + t^3 + t^3 + t^3) \\ & \quad \times (t^3 + t^3 + t^3 + t^3 + t^3 + t^3 + z_1^3 + z_2^3 + z_3^3) \\ &\ge (t^2 x_1 + t^2 x_2 + t^2 x_3 + t^2 y_1 + t^2 y_2 + t^2 y_3 + t^2 z_1 + t^2 z_2 + t^2 z_3)^3 \\ &= \frac{1}{36} \left( (x_1 + y_1 + z_1) + (x_2 + y_2 + z_2) + (x_3 + y_3 + z_3) \right)^3 \\ &\ge \frac{3^3}{36} (x_1 + y_1 + z_1)(x_2 + y_2 + z_2)(x_3 + y_3 + z_3). \end{align*}$$
Equality holds if and only if equality holds for each Cauchy-Schwarz and AM-GM, namely $x_1 = x_2 = x_3 = y_1 = y_2 = y_3 = z_1 = z_2 = z_3 = t$.

Therefore, the maximum value of $A$ is $\frac{3}{4}$ and equality holds if $x_1 = x_2 = x_3 = y_1 = y_2 = y_3 = z_1 = z_2 = z_3 = t$.