I finally figured out part of my second question, on whether this septic was a special case. The answer, perhaps not surprisingly, depends on Ramanujan's work. Given the Ramanujan theta function,

$$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$

Define the following theta quotients,

\begin{align} r_1 &= \frac{1}{q^{2/7}}\frac{f(-q^2,-q^5)}{f(-q,-q^6)} = \frac{1}{q^{2/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{7n-1})(1-q^{7n-6})}\\ r_2 &= \frac{-1\;}{q^{1/7}}\frac{f(-q^3,-q^4)}{f(-q^2,-q^5)} \;=\; \frac{-1\;}{q^{1/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-2})(1-q^{7n-5})}\\ r_3 &= \frac{1}{q^{-3/7}}\frac{f(-q,-q^6)}{f(-q^3,-q^7)} = \frac{1}{q^{-3/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{7n-3})(1-q^{7n-4})} \end{align}

Then the cubic formed by their $7$th powers

$$P(u) = (u-r_1^7)(u-r_2^7)(u-r_3^7) = 0$$

has coefficients in the Dedekind eta quotient $ m = \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4,$

$$P(u) = u^3- (57 + 14 m + m^2) u^2-(289 + 126 m + 19 m^2 + m^3) u +1 =0$$

while the septic formed by the expression,

$$P(y)=\prod_{k=0}^6 \Big(y - (\zeta^k r_1 + \zeta^{4k}r_2 + \zeta^{2k}r_3)\Big) = 0$$

with $\zeta = e^{2\pi i/7}$ also has coefficients in $m$,

$$P(y) = y^7 + 14y^4 + 7 (8 + m) y^3 + 14 (5 + m) y^2 - 28y - (113 + 21 m + m^2) = 0$$

Of course, a root of $P(y)$ is then,

$$y = r_1 + r_2 + r_3 = u_1^{1/7} + u_2^{1/7} + u_3^{1/7}$$

and a minor change of variable $m \to -(n+8)$ will recover the cubic and septic in my question.

P.S. While I now know how to construct the septic from first principles, I do not fully understand why its solvability "carries over" to any $m$ (not just the original eta quotient), nor why $P(y)$ now relates eta quotients in two ways,

$$\text{If}\; m = \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4,\; \text{then}\; y = \left(\frac{\eta(\tau/7)}{\eta(7\tau)}\right)+1$$

$$\text{If}\; m = \left(\frac{\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4,\; \text{then}\; y = \left(\frac{7\eta(49\tau)}{\eta(\tau)}\right)+1$$

I just figure Nature is very economical with her polynomials.

I finally figured out part of my second question, on whether this septic was a special case. The answer, perhaps not surprisingly, depends on Ramanujan's work. Given the Ramanujan theta function,

$$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$

Define the following theta quotients,

\begin{align} r_1 &= \frac{1}{q^{2/7}}\frac{f(-q^2,-q^5)}{f(-q,-q^6)} = \frac{1}{q^{2/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{7n-1})(1-q^{7n-6})}\\ r_2 &= \frac{-1\;}{q^{1/7}}\frac{f(-q^3,-q^4)}{f(-q^2,-q^5)} \;=\; \frac{-1\;}{q^{1/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-2})(1-q^{7n-5})}\\ r_3 &= \frac{1}{q^{-3/7}}\frac{f(-q,-q^6)}{f(-q^3,-q^7)} = \frac{1}{q^{-3/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{7n-3})(1-q^{7n-4})} \end{align}

Then the cubic formed by their $7$th powers

$$P(u) = (u-r_1^7)(u-r_2^7)(u-r_3^7) = 0$$

has coefficients in the Dedekind eta quotient $ m = \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4,$

$$P(u) = u^3- (57 + 14 m + m^2) u^2-(289 + 126 m + 19 m^2 + m^3) u +1 =0$$

(This cubic in fact was also found by Ramanujan.) While the septic formed by the expression,

$$P(y)=\prod_{k=0}^6 \Big(y - (\zeta^k r_1 + \zeta^{4k}r_2 + \zeta^{2k}r_3)\Big) = 0$$

with $\zeta = e^{2\pi i/7}$ also has coefficients in $m$,

$$P(y) = y^7 + 14y^4 + 7 (8 + m) y^3 + 14 (5 + m) y^2 - 28y - (113 + 21 m + m^2) = 0$$

Of course, a root of $P(y)$ is then,

$$y = r_1 + r_2 + r_3 = u_1^{1/7} + u_2^{1/7} + u_3^{1/7}$$

and a minor change of variable $m \to -(n+8)$ will recover the cubic and septic in my question.

P.S. While I now know how to construct the septic from first principles, I do not fully understand why its solvability "carries over" to any $m$ (not just the original eta quotient), nor why $P(y)$ now relates eta quotients in two ways,

$$\text{If}\; m = \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4,\; \text{then}\; y = \left(\frac{\eta(\tau/7)}{\eta(7\tau)}\right)+1$$

$$\text{If}\; m = \left(\frac{\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4,\; \text{then}\; y = \left(\frac{7\eta(49\tau)}{\eta(\tau)}\right)+1$$

I just figure Nature is very economical with her polynomials.