(This is not an answer, but a long comment to address the question by Somos.)

To find the powers of the $q$ factor, my method uses two parts: one is to use a formula by Ramanujan and second is to find an appropriate eta quotient. For example, for $p=13$ what I get is,

$$h_1 = \color{blue}{q^{179/156}}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$ $$h_2 = q^{119/156}\;\frac{f(-q^2,-q^{11})}{f(-q^5)}$$ $$h_3 = q^{35/156}\;\frac{f(-q^3,-q^{10})}{f(-q^5)}$$ $$h_4 = q^{-37/156}\;\frac{f(-q^4,-q^{9})}{f(-q^5)}$$ $$h_5 = q^{-61/156}\;\frac{f(-q^5,-q^{8})}{f(-q^5)}$$ $$h_6 = q^{-1/156}\;\frac{f(-q^6,-q^{7})}{f(-q^5)}$$

Their product is,

$$\prod_{k_1}^6 h_i = \frac{\eta(\tau)}{\eta(5\tau)}\left(\frac{\eta(13\tau)}{\eta(5\tau)}\right)^5$$

while their ratios have neat $q$-factor powers,

$$r_1 = \frac{h_2}{h_1} = q^{-5/13}\,F_1$$ $$r_2 = \frac{h_3}{h_2} = q^{-7/13}\,F_2$$ $$r_3 = \frac{h_4}{h_3} = q^{-6/13}\,F_3$$ $$r_4 = \frac{h_5}{h_4} = q^{-2/13}\,F_4$$ $$r_5 = \frac{h_6}{h_5} = q^{5/13}\,F_5$$ $$r_6 = \frac{h_1}{h_6} = q^{15/13}\,F_6$$

consistent with Ramanujan's ratio formula (disregarding signs) for $p=13$,

$$r_k = (-1)^{k-1}q^{k(3k-p)/(2p)}\,\frac{f(-q^{2k},-q^{p-2k})}{f(-q^{k},-q^{p-k})}$$

So I see nothing wrong with my,

$$h_1 = \color{blue}{q^{179/156}}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

while Somos has,

$$\text{Somos} = q^{149/156}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

But both cannot be radicals (since their ratio is just a power of $q$). P.S. I've re-done my calculations and $179$ is not a typo.

(This is not really an answer, but addresses the question by Somos. Caveat: It turns out his version of the six $q$ factors in the other answer are the correct ones.)

To find the powers of the $q$ factor, my method uses two parts: one is to use a formula by Ramanujan and second is to find an appropriate eta quotient. For example, for $p=13$, apparently,

$$h_1 = q^{179/156}\;\frac{f(-q,-q^{12})}{f(-q^5)}, \quad\quad h_2 = q^{119/156}\;\frac{f(-q^2,-q^{11})}{f(-q^5)}$$ $$h_3 = q^{35/156}\;\frac{f(-q^3,-q^{10})}{f(-q^5)},\quad\quad h_4 = q^{-37/156}\;\frac{f(-q^4,-q^{9})}{f(-q^5)}$$ $$h_5 = q^{-61/156}\;\frac{f(-q^5,-q^{8})}{f(-q^5)},\quad\quad h_6 = q^{-1/156}\;\frac{f(-q^6,-q^{7})}{f(-q^5)}$$

Their product is,

$$\prod_{k_1}^6 h_i = \frac{\eta(\tau)}{\eta(5\tau)}\left(\frac{\eta(13\tau)}{\eta(5\tau)}\right)^5$$

while their ratios have neat $q$-factor powers,

$$r_1 = \frac{h_2}{h_1} = q^{-5/13}\,F_1,\quad\quad r_2 = \frac{h_3}{h_2} = q^{-7/13}\,F_2$$ $$r_3 = \frac{h_4}{h_3} = q^{-6/13}\,F_3,\quad\quad r_4 = \frac{h_5}{h_4} = q^{-2/13}\,F_4$$ $$r_5 = \frac{h_6}{h_5} = q^{5/13}\,F_5,\quad\quad r_6 = \frac{h_1}{h_6} = q^{15/13}\,F_6$$

consistent with Ramanujan's ratio formula (disregarding signs) for $p=13$,

$$r_k = (-1)^{k-1}q^{k(3k-p)/(2p)}\,\frac{f(-q^{2k},-q^{p-2k})}{f(-q^{k},-q^{p-k})}$$

Edit:

It turns out that since levels $p=11, 13$ involves more functions than $p=7$, I inadvertently used the wrong pairs to derive the $q$ factor. Thus, the correct version is by Somos,

$$\text{Somos} = q^{149/156}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

The product given above remains accurate though.

(This is not an answer, but a long comment to address the question by Somos.)

To find the powers of the $q$ factor, my method uses two parts: one is to use a formula by Ramanujan and second is to find an appropriate eta quotient. For example, for $p=13$ what I get is,

$$h_1 = \color{blue}{q^{179/156}}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$ $$h_2 = q^{119/156}\;\frac{f(-q^2,-q^{11})}{f(-q^5)}$$ $$h_3 = q^{35/156}\;\frac{f(-q^3,-q^{10})}{f(-q^5)}$$ $$h_4 = q^{-37/156}\;\frac{f(-q^4,-q^{9})}{f(-q^5)}$$ $$h_5 = q^{-61/156}\;\frac{f(-q^5,-q^{8})}{f(-q^5)}$$ $$h_6 = q^{-1/156}\;\frac{f(-q^6,-q^{7})}{f(-q^5)}$$

Their product is,

$$\prod_{k_1}^6 h_i = \frac{\eta(\tau)}{\eta(5\tau)}\left(\frac{\eta(13\tau)}{\eta(5\tau)}\right)^5$$

while their ratios have neat $q$-factor powers,

$$r_1 = \frac{h_2}{h_1} = q^{-5/13}\,F_1$$ $$r_2 = \frac{h_3}{h_2} = q^{-7/13}\,F_2$$ $$r_3 = \frac{h_4}{h_3} = q^{-6/13}\,F_3$$ $$r_4 = \frac{h_5}{h_4} = q^{-2/13}\,F_4$$ $$r_5 = \frac{h_6}{h_5} = q^{5/13}\,F_5$$ $$r_6 = \frac{h_1}{h_6} = q^{15/13}\,F_6$$

consistent with Ramanujan's ratio formula (disregarding signs) for $p=13$,

$$r_k = (-1)^{k-1}q^{k(3k-p)/(2p)}\,\frac{f(-q^{2k},-q^{p-2k})}{f(-q^{k},-q^{p-k})}$$

So I see nothing wrong with my,

$$h_1 = \color{blue}{q^{179/156}}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

while Somos has,

$$\text{Somos} = q^{199/156}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

But both cannot be radicals (since their ratio is just a power of $q$). P.S. I've re-done my calculations and $179$ is not a typo.

(This is not an answer, but a long comment to address the question by Somos.)

To find the powers of the $q$ factor, my method uses two parts: one is to use a formula by Ramanujan and second is to find an appropriate eta quotient. For example, for $p=13$ what I get is,

$$h_1 = \color{blue}{q^{179/156}}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$ $$h_2 = q^{119/156}\;\frac{f(-q^2,-q^{11})}{f(-q^5)}$$ $$h_3 = q^{35/156}\;\frac{f(-q^3,-q^{10})}{f(-q^5)}$$ $$h_4 = q^{-37/156}\;\frac{f(-q^4,-q^{9})}{f(-q^5)}$$ $$h_5 = q^{-61/156}\;\frac{f(-q^5,-q^{8})}{f(-q^5)}$$ $$h_6 = q^{-1/156}\;\frac{f(-q^6,-q^{7})}{f(-q^5)}$$

Their product is,

$$\prod_{k_1}^6 h_i = \frac{\eta(\tau)}{\eta(5\tau)}\left(\frac{\eta(13\tau)}{\eta(5\tau)}\right)^5$$

while their ratios have neat $q$-factor powers,

$$r_1 = \frac{h_2}{h_1} = q^{-5/13}\,F_1$$ $$r_2 = \frac{h_3}{h_2} = q^{-7/13}\,F_2$$ $$r_3 = \frac{h_4}{h_3} = q^{-6/13}\,F_3$$ $$r_4 = \frac{h_5}{h_4} = q^{-2/13}\,F_4$$ $$r_5 = \frac{h_6}{h_5} = q^{5/13}\,F_5$$ $$r_6 = \frac{h_1}{h_6} = q^{15/13}\,F_6$$

consistent with Ramanujan's ratio formula (disregarding signs) for $p=13$,

$$r_k = (-1)^{k-1}q^{k(3k-p)/(2p)}\,\frac{f(-q^{2k},-q^{p-2k})}{f(-q^{k},-q^{p-k})}$$

So I see nothing wrong with my,

$$h_1 = \color{blue}{q^{179/156}}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

while Somos has,

$$\text{Somos} = q^{149/156}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

But both cannot be radicals (since their ratio is just a power of $q$). P.S. I've re-done my calculations and $179$ is not a typo.