I discovered the following formula while working on the sequence A141697 from the OEIS. I have no idea whether it is something trivial or not. I would be very happy to know more about it.
$$ \textrm{Li}_\nu\left(z\right)= \frac { 6 (1+z)^{-\nu-1} + \displaystyle\sum_{k=0}^{-\nu-1} \displaystyle\left( -6 \displaystyle{{-\nu-1}\displaystyle\choose k}+7\sum_{j=0}^{k+1}(-1)^j (k-j+1)^{-\nu} {{-\nu+1}\ \choose j} \right) z^k } { 7 (1 - z)^{-\nu+1} }z $$ when $\nu$ is a negative integer.
In case I mistyped the formula, here are my codes in two different languages:
For Pari-GP:
mypolylog(n, x) = { ( 6*(x+1)^(-n-1)
+ sum(k=0,-n-1, (-6*binomial(-n-1,k)
+ 7*sum(j=0,k+1, (-1)^j * (k-j+1)^(-n) * binomial(-n+1,j)))*x^k) ) * x
/ (7*(1-x)^(-n+1) ) }
For Mathematica:
mypolylog[n_, x_] := (6*(x+1)^(-n-1)
+ Sum[(x^k*(-6*Binomial[-n-1, k]
+ 7*Sum[(-1)^j*(k-j+1)^(-n)*Binomial[-n+1, j], {j, 0, k+1}])),
{k, 0, -n-1}]) / ( 7*(1 - x)^(-n+1) ) * x
To simplify formulas define $\, B(n,k) := {-n-1\choose k}, \,$ $\, A(n,k) := \sum_{j=0}^{k+1} (-1)^j (k-j+1)^{-n} {-n+1 \choose j}, \,$ and $\, u := 1-t \,$ where $\,n<0\,$ is a negative integer and $\, t\neq 0. \,$ Your formula is the $\, t=7 \,$ case of the lightly simplified equation
$$ \textrm{Li}_n(z) = \frac{z}t (1-z)^{n-1} (-u (1-z)^{-n+1} + u \sum_{k=0}^{-n-1} z^k B(n,k) + t \sum_{k=0}^{-n-1} z^k A(n,k) ). $$ But $\, (1-z)^{-n+1} = \sum_{k=0}^{-n-1} z^k B(n,k) \,$ simplies it to $ \textrm{Li}_n(z) = z (1-z)^{n-1} \sum_{k=0}^{-n-1} z^k A(n,k). \,$
The numbers in $\,A(-n,k-1)\,$ are the triangular OEIS sequence A008292 of Eulerian numbers which entry has the information "O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x".