If one of $a, b, c, d$ is $0$, it is easy to deduce that $a = b = c = d = 0$ is a solution, so in what follows we may assume $a, b, c, d > 0$.
After rearranging the original equations, we obtain
$$a^{12}(b+c)^3(b+c+d)^4 = 1$$
$$b^{12}(c+d)^3(c+d+a)^4 = 1$$
$$c^{12}(d+a)^3(d+a+b)^4 = 1$$
$$d^{12}(a+b)^3(a+b+c)^4 = 1$$
Consider $x$ to be the largest among the $16$ numbers $\left\{\frac{x}{y}\mid x, y \in a, b, c, d\right\}$, and without loss of generality assume $\frac{a}{b} = x \ge 1$. Then
$$\frac{c+d}{b+c} \le x$$
$$\frac{c+d+a}{b+c+d} \le x$$
Then
$$1 = \frac{a^{12}(b+c)^3(b+c+d)^4}{b^{12}(c+d)^3(c+d+a)^4} \ge x^{12} \frac{1}{x^3} \frac{1}{x^4} = x^7$$
Therefore $x = 1$, that is $a = b = c = d$.

(Note: In the conditions given by the problem, $a, b, c, d$ do not have full symmetry, but only a cyclic relationship, so the assumption that is truly without loss of generality should consider the three cases $x = \frac{a}{b}, \frac{a}{c}, \frac{a}{d}$, but the above derivation still applies to all of them.)

Substituting back into the original equations, we solve to get $(a, b, c, d) = (0, 0, 0, 0)$ or $\left(\frac{1}{\sqrt{19/648}}, \frac{1}{\sqrt{19/648}}, \frac{1}{\sqrt{19/648}}, \frac{1}{\sqrt{19/648}}\right)$