By the symmetry of the given conditions, we may assume without loss of generality that $a$ is the largest and $d$ is the smallest. From the assumption we know $a \ge 0$, $d \le 0$ and $d = -(a+b+c) \le 0$, that is, $a+b+c \ge 0$.
Let $S_k = a^k + b^k + c^k + d^k$, where $k$ is a positive integer, then $S_7 = a^7 + b^7 + c^7 - (a+b+c)^7$.
Because when $a = -b$ or $b = -c$ or $c = -a$, $S_7 = 0$. Hence we may assume
$$S_7 = (a+b)(b+c)(c+a)[x(a^4 + b^4 + c^4) + y(a^3b + ab^3 + b^3c + bc^3 + c^3a + ca^3) + z(a^2b^2 + b^2c^2 + c^2a^2) + wabc(a+b+c)].$$
(Because $S_7$ is a symmetric homogeneous polynomial in $a, b, c$.)
In the above expression, taking respectively $a = b = 1, c = 0$; $a = b = c = 1$; $a = b = 1, c = 2$; $a = b = 1, c = 3$, we obtain
$$\begin{cases} 2x + 2y + z + 63 = 0, \\ x + 2y + z + w + 91 = 0, \\ 54x + 66y + 27z + 24w + 2709 = 0, \\ 332x + 248y + 76z + 60w + 9492 = 0. \end{cases}$$

Solving, we get $x = -7, y = -14, z = -21, w = -35$. Hence, we obtain
$$\begin{align*} S_7 &= -7(a+b)(b+c)(c+a)[(a^4 + b^4 + c^4) \\ &\quad +2(a^3b + ab^3 + b^3c + bc^3 + c^3a + ca^3) \\ &\quad +3(a^2b^2 + b^2c^2 + c^2a^2) + 5abc(a+b+c)] \\ &= -7(a+b)(b+c)(c+a)[(a^2 + b^2 + c^2)^2 \\ &\quad +2(ab + bc + ca)(a^2 + b^2 + c^2) + (ab + bc + ca)^2 + abc(a+b+c)] \\ &= -7(a+b)(b+c)(c+a)[(a^2 + b^2 + c^2 + ab + bc + ca)^2 + abc(a+b+c)]. \end{align*}$$
$$\begin{align*} S_2 &= a^2 + b^2 + c^2 + (a+b+c)^2 \\ &= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\ &= (a+b)^2 + (b+c)^2 + (c+a)^2. \end{align*}$$
By the AM-GM inequality, we get
$$27(a + b)^2(b + c)^2(c + a)^2 \\ \le [(a + b)^2 + (b + c)^2 + (c + a)^2]^3.$$
Also, from
$$abc(a + b + c) \le \frac{1}{3}(ab + bc + ca)^2, \ ab + bc + ca \le a^2 + b^2 + c^2,$$
we get
$$\begin{align*} & 48[(a^2 + b^2 + c^2 + ab + bc + ca)^2 + abc(a + b + c)] \\ & \le 48(a^2 + b^2 + c^2 + ab + bc + ca)^2 + 16(ab + bc + ca)^2 \\ & \le 52(a^2 + b^2 + c^2 + ab + bc + ca)^2 \\ & = 13[(a + b)^2 + (b + c)^2 + (c + a)^2]^2. \end{align*}$$

1296S_7^2
$$$$
$$\begin{align*} &= 49[27(a+b)^2(b+c)^2(c+a)^2] \times \\ &\quad \{48[(a^2+b^2+c^2+ab+bc+ca)^2+abc(a+b+c)]\}^2 \\ &\le 49[(a+b)^2+(b+c)^2+(c+1)^2]^3 \{13[(a+b)^2+(b+c)^2+(c+a)^2]^2\}^2 \\ &= 637[(a+b)^2+(b+c)^2+(c+a)^2]^7. \end{align*}$$
$$$$