Multiplying all equations gives
$$\left(a^2 + \frac{1}{b^2}\right) \left(b^2 + \frac{4}{c^2}\right) \left(c^2 + \frac{16}{d^2}\right) \left(d^2 + \frac{4}{a^2}\right) = 2^8.$$
By AM-GM, $a^2 + \frac{1}{b^2} \ge 2 \cdot \frac{a}{b}$, where the equality holds if and only if $a = \frac{1}{b}$. Similarly $b^2 + \frac{4}{c^2} \ge 4 \cdot \frac{b}{c}$ (equality if and only if $b = \frac{2}{c}$), $c^2 + \frac{16}{d^2} \ge 8 \cdot \frac{c}{d}$ (equality if and only if $c = \frac{4}{d}$) and $d^2 + \frac{4}{a^2} \ge 4 \cdot \frac{d}{a}$ (equality if and only if $d = \frac{2}{a}$). Multiplying the obtained four inequalities gives
$$\left(a^2 + \frac{1}{b^2}\right) \left(b^2 + \frac{4}{c^2}\right) \left(c^2 + \frac{16}{d^2}\right) \left(d^2 + \frac{4}{a^2}\right) \ge 2 \cdot \frac{a}{b} \cdot 4 \cdot \frac{b}{c} \cdot 8 \cdot \frac{c}{d} \cdot 4 \cdot \frac{d}{a} = 2^8.$$
The resulting inequality must hold as equality by the first step of the solution. This is possible only if all four inequalities hold as equalities, whence
$$\begin{cases} a = \frac{1}{b}, \\ b = \frac{2}{c}, \\ c = \frac{4}{d}, \\ d = \frac{2}{a}. \end{cases}$$
By multiplying the equations of this system, we get $abcd = \frac{16}{abcd}$, whence $(abcd)^2 = 16$. As $a, b, c$ and $d$ are positive, the only possibility is $abcd = 4$.

By introducing $a = \frac{x}{2}$, $b = 2y$, $c = z$, $d = 4t$, rewrite the system as
$$\begin{cases} \frac{x^2}{4} + \frac{1}{4y^2} = \frac{1}{2}, \\ 4y^2 + \frac{4}{z^2} = 8, \\ z^2 + \frac{16}{16t^2} = 2, \\ 16t^2 + \frac{4 \cdot 4}{x^2} = 32. \end{cases}$$
Multiplying the first equation by 4, the second equation by $\frac{1}{4}$, and the fourth equation by $\frac{1}{16}$, we obtain an equivalent system
$$\begin{cases} x^2 + \frac{1}{y^2} = 2, \\ y^2 + \frac{1}{z^2} = 2, \\ z^2 + \frac{1}{t^2} = 2, \\ t^2 + \frac{1}{x^2} = 2. \end{cases}$$
Adding the equations of this system gives
$$x^2 + \frac{1}{y^2} + y^2 + \frac{1}{z^2} + z^2 + \frac{1}{t^2} + t^2 + \frac{1}{x^2} = 8.$$
But for every real number $u$, $u + \frac{1}{u} \ge 2$, where equality holds only if $u = 1$. By adding up inequalities $x^2 + \frac{1}{x^2} \ge 2$, $y^2 + \frac{1}{y^2} \ge 2$, $z^2 + \frac{1}{z^2} \ge 2$ and $t^2 + \frac{1}{t^2} \ge 2$, we get
$$x^2 + \frac{1}{x^2} + y^2 + \frac{1}{y^2} + z^2 + \frac{1}{z^2} + t^2 + \frac{1}{t^2} \ge 8.$$
This inequality must hold as equality by the above; hence all four previous inequalities must hold as equalities, too, i.e., $x^2 = y^2 = z^2 = t^2 = 1$. As the numbers are positive, the only possibility is $x = y = z = t = 1$. Hence $abcd = \frac{x}{2} \cdot 2y \cdot z \cdot 4t = 4xyzt = 4$.