Being: $$\sum_{n=-\infty}^{\infty}(-1)^nq^{n(3n-1)/2}(3n^2-n) \tag{1},$$ it can be shown: $$\sum_{n=-\infty}^{\infty}(-1)^nq^{n(3n-1)/2}(3n^2-n)=-\frac{1}{12}(1-P(q))\prod_{n=1}^{\infty}(1-q^n)\tag{2},$$ where $$P(q)=1-24\sum_{n=1}^{\infty}\frac{nq^{n}}{1-q^{n}}\tag{3}.$$
Then beautiful series like: $$\sum_{n=-\infty}^{\infty}(-1)^ne^{-\pi n(3n-1)/2}(3n^2-n)=2^{1/8}e^{\pi/24}\left(\frac{\Gamma{(\frac{1}{4})}(6-\pi)}{24\pi^{7/4}}-\frac{\Gamma{(\frac{1}{4})}^5}{64\pi^{15/4}} \right)\tag{4},$$ can be obtained.
Question Can we improve this result to the consecutive derivatives of $(2)$ and get a more general result?
Ramanujan dealt with the functions $P, Q, R$ and their derivatives in great detail in his paper On certain arithmetical functions using elementary algebra.
We have the definitions \begin{align} \eta(q) &:= q^{1/24}\prod_{n=1}^{\infty}(1-q^n)\tag{1}\\ P(q) &:= 1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}\tag{2}\\ Q(q) &:= 1+240\sum_{n=1}^{\infty}\frac{n^3q^n}{1-q^n}\tag{3}\\ R(q) &:= 1-504\sum_{n=1}^{\infty}\frac{n^5q^n}{1-q^n}\tag{4}\\ \end{align}
Euler's pentagonal theorem says that $$\Phi(q) :=q^{-1/24}\eta(q)=\prod_{n=1}^{\infty}(1-q^n)=\sum_{n\in\mathbb{Z}}(-1)^nq^{n(3n-1)/2}\tag{5}$$ Let us also note that $P(q) $ is nothing but logarithmic derivative of $\eta(q) $ ie $$P(q)=24q\frac{d}{dq}\log\eta(q)\tag{6}$$ The sum in question, say $F(q)$, is defined as $$F(q) =\sum_{n\in\mathbb{Z}} (-1)^n(3n^2-n)q^{n(3n-1)/2}$$ and this clearly equals $$2q\frac{d\Phi(q)}{dq}=2q\frac{d}{dq}(q^{-1/24}\eta(q)) =-\frac{q^{-1/24}\eta(q)}{12}+2q^{-1/24}\eta(q)q\frac{d}{dq}\log\eta(q)=\frac{\Phi(q)}{12}(P(q)-1)$$ so that the second equation in question is verified.
The values of $\eta(q), P(q) $ can be evaluated in closed form in terms of $\pi$ and values of gamma function at rational points if $q=\exp(-\pi\sqrt{r}) $ and $r$ is a positive rational number and some of these values are well known.
The evaluations are effected by using their link with elliptic integrals. We introduce them with a brief summary. Let $k\in(0,1)$ be elliptic modulus and $k'=\sqrt{1-k^2}$ be the complementary (to $k$) modulus and we define elliptic integral of first kind $$K(k) =\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{7}$$ and elliptic integral of second kind $$E(k) =\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx\tag{8}$$ The integrals $K(k), E(k), K(k'), E(k') $ are usually denoted by $K, E, K', E'$ if the moduli $k, k'$ are available from context and they satisfy a relation $$KE'+K'E-KK'=\frac{\pi} {2}\tag{9}$$ which goes by the name Legendre's identity. The values of $k, k'$ can be obtained from the values of $K, K'$ using a variable $q=\exp(-\pi K'/K) $ called nome corresponding to $q$. This happens using the Jacobi theta functions \begin{align} \vartheta_2(q)&:=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}=2q^{1/4}\prod_{n=1}^{\infty}(1-q^{2n})(1+q^{2n})^2\tag{10}\\ \vartheta_3(q)&:=\sum_{n\in\mathbb{Z}}q^{n^2}=\prod_{n=1}^{\infty}(1-q^{2n})(1+q^{2n-1})^2\tag{11}\\ \vartheta_4(q)&:=\vartheta_3(-q)=\prod_{n=1}^{\infty}(1-q^{2n})(1-q^{2n-1})^2\tag{12} \end{align} The product representations above are an immediate consequence of Jacobi triple product identity $$\sum_{n\in\mathbb{Z}} z^nq^{n^2}=\prod_{n=1}^{\infty}(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1}),\,z\neq 0\tag{13}$$ The link between theta functions and elliptic integrals is given by the formulas $$k=\frac{\vartheta_2^2(q)}{\vartheta_3^2(q)}, k'=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)},\frac{2K}{\pi}=\vartheta_3^2(q)\tag{14}$$
The most well known evaluation of elliptic integrals and theta functions is done when $q=e^{-\pi} $ so that $K'=K$ and $k=k'=1/\sqrt{2}$ and the elliptic integral $K$ can be evaluated with help of beta function as $K=\Gamma^2(1/4)/(4\sqrt{\pi})$. The evaluation of $\eta(q) $ is possible once we observe that $$\eta(q) =2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{15}$$ via product representation of $\eta(q), k, k', K$. The evaluation of $P(q) $ is possible by noting that it is the logarithmic derivative of $\eta(q) $ and we have the relation $$P(q) =\left(\frac{2K}{\pi}\right) ^2\left(\frac{6E}{K}+k^2-5\right)\tag{16}$$ The above formula is obtained by differentiating $(15)$ with respect to $q$ and noting that $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2},\frac{dK}{dk}=\frac{E-k'^2K}{kk'^2}\tag{17}$$ Using $(15),(16)$ we can evaluate the function given in question ($F(q) $ of this answer) in closed form.
To obtain higher derivatives of $\eta(q) $ we need to differentiate $P(q), Q(q), R(q) $ and Ramanujan proved (in the paper mentioned at the beginning of the answer) that \begin{align} q\frac{dP} {dq} &=\frac{P^2-Q}{12}\tag{18}\\ q\frac{dQ}{dq}&=\frac{PQ-R}{3}\tag{19}\\ q\frac{dR}{dq}&=\frac{PR-Q^2}{2}\tag{20} \end{align} A general formula for the repeated derivates of these functions is not known and hence the calculations must be performed as and when needed.
Further note that we also need evaluations of the functions $Q, R$ and the above derivative formulas can be used to prove that $$Q(q) =\left(\frac{2K}{\pi}\right) ^4(1+14k^2+k^4)\tag{21}$$ and $$R(q) =\left(\frac{2K}{\pi}\right) ^6(1+k^2)(1-34k^2+k^4)\tag{22}$$ which help us to evaluate $Q, R$ in closed form.