Very interesting discussion, and I would like to put there my version of this problem.

First, all identities only have any sense if basic conventional points are postulated. Namely, below I will use the following definition of four-dimensional Levi-Civita symbol: $$ \varepsilon^{\alpha\beta\gamma\delta}=-\varepsilon_{\alpha\beta\gamma\delta}= \begin{cases} +1,& \text{if } \alpha\beta\gamma\delta=\text{even permutation of 0123}\\ -1,& \text{if } \alpha\beta\gamma\delta=\text{odd permutation of 0123}. \end{cases} $$ Next, the definition of fifth gamma-matrix is the following: $$ \gamma^5=\gamma_5=i\gamma^0\gamma^1\gamma^2\gamma^3=-i\gamma_0\gamma_1\gamma_2\gamma_3. $$ With such definitions $$ \mathrm{Tr}\{{\gamma^5\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}}\}=\mathrm{Tr}\{{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}\gamma^5}\}=-4i\varepsilon^{\alpha\beta\gamma\delta},\tag{1}\label{1} $$ what is checked for the component $\alpha=0,\ \beta=1,\ \gamma=2,\ \delta=3$ and is carried over to the other components which are permutations of 0123.

Further, there is the identity $$ \gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}=\eta^{\alpha\beta}\gamma^{\gamma}-\eta^{\alpha\gamma}\gamma^{\beta}+\eta^{\beta\gamma}\gamma^{\alpha}-i\gamma^5\varepsilon^{\alpha\beta\gamma\delta}\gamma_{\delta},\tag{2}\label{2} $$ where $\eta^{\mu\nu}$ is the Minkowski metric with the signature $\{+1, -1, -1, -1\}$ and, hence, $\gamma_0=\gamma^0$, while $\gamma_\mu=-\gamma^\mu$ with $\mu=\{1, 2, 3\}$. Such defined the identity is in agreement with the version of @J.G. because $\gamma^5$ anticommutes with any of $\gamma_\mu$ and $\varepsilon^{\delta\alpha\beta\gamma}=-\varepsilon^{\alpha\beta\gamma\delta}$.

Finally, we are ready to calculate the trace $\mathrm{Tr}\{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\rho}\gamma^{\sigma}\gamma^{\tau}\gamma^5\}$.

First, replacing $\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}$ with the use of the identity \ref{2} one obtains $$ \mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}= \eta^{\alpha\beta}\mathrm{Tr}\{\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}- \eta^{\alpha\gamma}\mathrm{Tr}\{\gamma^\beta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}+ \eta^{\beta\gamma}\mathrm{Tr}\{\gamma^\alpha\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}- i\varepsilon^{\alpha\beta\gamma\delta}\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}.\tag{3}\label{3} $$ First three terms are rewritten with the use of the identity \ref{1}. As for the last term, let's consider the multiplier $\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}$ separately: $$ \mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\} $$ because of anticommutation of $\gamma^5$ with both $\gamma_\mu$ and $\gamma^\mu$, further use the identity \ref{2} one more time: $$ \mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}= \eta^{\rho\sigma}\mathrm{Tr}\{\gamma_\delta\gamma^\tau\}- \eta^{\rho\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\sigma\}+ \eta^{\sigma\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\rho\}- i\varepsilon^{\rho\sigma\tau\eta}\mathrm{Tr}\{\gamma_\delta\gamma^5\gamma_\eta\}.\tag{4}\label{4} $$ The last term vanishes (trace of product of fifth gamma-matrix only two gamma-matrices) and every of the other terms contains the multiplier of the form $$ \mathrm{Tr}\{\gamma_\lambda\gamma^\nu\}=\eta_{\lambda\mu}\mathrm{Tr}\{\gamma^\mu\gamma^\nu\}=4\eta_{\lambda\mu}\eta^{\mu\nu}. $$ Using this one can rewrite \ref{4} in the form $$ \mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}= 4\eta^{\rho\sigma}\eta_{\delta\lambda}\eta^{\lambda\tau}- 4\eta^{\rho\tau}\eta_{\delta\lambda}\eta^{\lambda\sigma}+ 4\eta^{\sigma\tau}\eta_{\delta\lambda}\eta^{\lambda\rho} $$ and inserting it to \ref{3} one obtains $$ \mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=-4i( \eta^{\alpha\beta}\varepsilon^{\gamma\rho\sigma\tau} -\eta^{\alpha\gamma}\varepsilon^{\beta\rho\sigma\tau} +\eta^{\beta\gamma}\varepsilon^{\alpha\rho\sigma\tau} +\eta^{\rho\sigma}\varepsilon^{\alpha\beta\gamma\tau} -\eta^{\rho\tau}\varepsilon^{\alpha\beta\gamma\sigma} +\eta^{\sigma\tau}\varepsilon^{\alpha\beta\gamma\rho}). $$ Note, that we have used the following transformations: $$ \varepsilon^{\alpha\beta\gamma\delta}\eta_{\delta\lambda}\eta^{\lambda\tau}= \varepsilon^{\alpha\beta\gamma}_{\phantom{\alpha\beta\gamma}\lambda}\eta^{\lambda\tau}= \varepsilon^{\alpha\beta\gamma\tau} $$ and so on.

Very interesting discussion, and I would like to put there my version of this problem.

First, all identities only have any sense if basic conventional points are postulated. Namely, below I will use the following definition of four-dimensional Levi-Civita symbol: $$ \varepsilon^{\alpha\beta\gamma\delta}=-\varepsilon_{\alpha\beta\gamma\delta}= \begin{cases} +1,& \text{if } \alpha\beta\gamma\delta=\text{even permutation of 0123}\\ -1,& \text{if } \alpha\beta\gamma\delta=\text{odd permutation of 0123}. \end{cases} $$ Next, the definition of fifth gamma-matrix is the following: $$ \gamma^5=\gamma_5=i\gamma^0\gamma^1\gamma^2\gamma^3=-i\gamma_0\gamma_1\gamma_2\gamma_3. $$ With such definitions $$ \mathrm{Tr}\{{\gamma^5\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}}\}=\mathrm{Tr}\{{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}\gamma^5}\}=-4i\varepsilon^{\alpha\beta\gamma\delta},\tag{1}\label{1} $$ what is checked for the component $\alpha=0,\ \beta=1,\ \gamma=2,\ \delta=3$ and is carried over to the other components which are permutations of 0123.

Further, there is the identity $$ \gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}=\eta^{\alpha\beta}\gamma^{\gamma}-\eta^{\alpha\gamma}\gamma^{\beta}+\eta^{\beta\gamma}\gamma^{\alpha}-i\gamma^5\varepsilon^{\alpha\beta\gamma\delta}\gamma_{\delta},\tag{2}\label{2} $$ where $\eta^{\mu\nu}$ is the Minkowski metric with the signature $\{+1, -1, -1, -1\}$ and, hence, $\gamma_0=\gamma^0$, while $\gamma_\mu=-\gamma^\mu$ with $\mu=\{1, 2, 3\}$. Such defined the identity is in agreement with the version of @J.G. because $\gamma^5$ anticommutes with any of $\gamma_\mu$ and $\varepsilon^{\delta\alpha\beta\gamma}=-\varepsilon^{\alpha\beta\gamma\delta}$.

Finally, we are ready to calculate the trace $\mathrm{Tr}\{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\rho}\gamma^{\sigma}\gamma^{\tau}\gamma^5\}$.

First, replacing $\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}$ with the use of the identity \ref{2} one obtains $$ \mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}= \eta^{\alpha\beta}\mathrm{Tr}\{\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}- \eta^{\alpha\gamma}\mathrm{Tr}\{\gamma^\beta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}+ \eta^{\beta\gamma}\mathrm{Tr}\{\gamma^\alpha\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}- i\varepsilon^{\alpha\beta\gamma\delta}\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}.\tag{3}\label{3} $$ First three terms are rewritten with the use of the identity \ref{1}. As for the last term, let's consider the multiplier $\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}$ separately: $$ \mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\} $$ because of anticommutation of $\gamma^5$ with both $\gamma_\mu$ and $\gamma^\mu$, further use the identity \ref{2} one more time: $$ \mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}= \eta^{\rho\sigma}\mathrm{Tr}\{\gamma_\delta\gamma^\tau\}- \eta^{\rho\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\sigma\}+ \eta^{\sigma\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\rho\}- i\varepsilon^{\rho\sigma\tau\eta}\mathrm{Tr}\{\gamma_\delta\gamma^5\gamma_\eta\}.\tag{4}\label{4} $$ The last term vanishes (trace of product of fifth gamma-matrix with only two gamma-matrices) and every of the other terms contains the multiplier of the form $$ \mathrm{Tr}\{\gamma_\lambda\gamma^\nu\}=\eta_{\lambda\mu}\mathrm{Tr}\{\gamma^\mu\gamma^\nu\}=4\eta_{\lambda\mu}\eta^{\mu\nu}. $$ Using this one can rewrite \ref{4} in the form $$ \mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}= 4\eta^{\rho\sigma}\eta_{\delta\lambda}\eta^{\lambda\tau}- 4\eta^{\rho\tau}\eta_{\delta\lambda}\eta^{\lambda\sigma}+ 4\eta^{\sigma\tau}\eta_{\delta\lambda}\eta^{\lambda\rho} $$ and inserting it to \ref{3} one obtains $$ \mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=-4i( \eta^{\alpha\beta}\varepsilon^{\gamma\rho\sigma\tau} -\eta^{\alpha\gamma}\varepsilon^{\beta\rho\sigma\tau} +\eta^{\beta\gamma}\varepsilon^{\alpha\rho\sigma\tau} +\eta^{\rho\sigma}\varepsilon^{\alpha\beta\gamma\tau} -\eta^{\rho\tau}\varepsilon^{\alpha\beta\gamma\sigma} +\eta^{\sigma\tau}\varepsilon^{\alpha\beta\gamma\rho}). $$ Note, that we have used the following transformations: $$ \varepsilon^{\alpha\beta\gamma\delta}\eta_{\delta\lambda}\eta^{\lambda\tau}= \varepsilon^{\alpha\beta\gamma}_{\phantom{\alpha\beta\gamma}\lambda}\eta^{\lambda\tau}= \varepsilon^{\alpha\beta\gamma\tau} $$ and so on.