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(Updated answer):

Upon further research, it turns out your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.

I. Ladder of index 12

$$A = \frac{\text{Li}_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$

$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$

then for $n=1,2,3,4,5$,

$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$

with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation.

II. Ladder of index 20

Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.

III. Ladder of index 24

Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.

(Old answer): As to your question if there are similar identities, it seems there is a whole family,

$$5\,\text{Li}_2(\phi^{-1}) =-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$

$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$

$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$

$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ =-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$

$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ =m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$

$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ =n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$

and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here.

Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$.

P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.

(Updated answer):

Upon further research, it turns out your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.

I. Ladder of index 12

$$A = \frac{\text{Li}_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$

$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$

then for $n=1,2,3,4,5$,

$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$

with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation. Incidentally, a consequence (or is it the other way round?) of this ladder is the high-degree cyclotomic equation satisfied by the golden ratio,

$$(1 - \phi^{-12}) = (1 - \phi^{-6})^{3/2}(1 - \phi^{-4})(1 - \phi^{-2})^{-11/48}\,\phi^{13/48}$$

Notice that the powers of $(1 - \phi^{k})$ are coefficients in $A,B$.

II. Ladder of index 20

Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.

III. Ladder of index 24

Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.

(Old answer): As to your question if there are similar identities, it seems there is a whole family,

$$5\,\text{Li}_2(\phi^{-1}) =-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$

$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$

$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$

$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ =-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$

$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ =m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$

$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ =n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$

and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here.

Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$.

P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.

(Updated answer):

Your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.

I. Ladder of index 12

$$A = \frac{\text{Li}_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$

$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$

then for $n=1,2,3,4,5$,

$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$

with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation.

II. Ladder of index 20

Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.

III. Ladder of index 24

Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.

(Old answer): As to your question if there are similar identities, it seems there is a whole family,

$$5\,\text{Li}_2(\phi^{-1}) =-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$

$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$

$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$

$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ =-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$

$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ =m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$

$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ =n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$

and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here.

Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$.

P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.

(Updated answer):

Upon further research, it turns out your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.

I. Ladder of index 12

$$A = \frac{\text{Li}_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$

$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$

then for $n=1,2,3,4,5$,

$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$

with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation.

II. Ladder of index 20

Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.

III. Ladder of index 24

Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.

(Old answer): As to your question if there are similar identities, it seems there is a whole family,

$$5\,\text{Li}_2(\phi^{-1}) =-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$

$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$

$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$

$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ =-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$

$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ =m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$

$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ =n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$

and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here.

Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$.

P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.

(Updated answer):

Your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.

I. Ladder of index 12

$$A = \frac{\text{Li}_n(\phi^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$

$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$

then for $n=1,2,3,4,5$,

$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$

with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation.

II. Ladder of index 20

Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.

III. Ladder of index 24

Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.

(Old answer): As to your question if there are similar identities, it seems there is a whole family,

$$5\,\text{Li}_2(\phi^{-1}) =-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$

$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$

$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$

$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ =-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$

$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ =m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$

$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ =n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$

and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here.

Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$.

P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.

(Updated answer):

Your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.

I. Ladder of index 12

$$A = \frac{\text{Li}_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$

$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$

then for $n=1,2,3,4,5$,

$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$

with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation.

II. Ladder of index 20

Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.

III. Ladder of index 24

Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.

(Old answer): As to your question if there are similar identities, it seems there is a whole family,

$$5\,\text{Li}_2(\phi^{-1}) =-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$

$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$

$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$

$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ =-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$

$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ =m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$

$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ =n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$

and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here.

Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$.

P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.