$$\sum_{n=1}^\infty\frac{\cot(\pi n\sqrt{61})}{n^3}=-\frac{16793\pi^3}{45660\sqrt{61}}.$$
Prove it converges and,
evaluate the series.
For the first part of the question, I prove it converges by considering the irrationality measure,
$$|\sin (n\pi\sqrt{61})|=|\sin (n\pi\sqrt{61}-m\pi)|\ge \frac{2}{\pi}|n\pi\sqrt{61}-m\pi|\ge 2n\left|\sqrt{61}-\frac{m}{n}\right|>\frac{C}{n},$$
so $$\left|\frac{\cot(\pi n\sqrt{61})}{n^3}\right|\le\left|\frac{1}{n^3\sin(\pi n\sqrt{61})}\right|<\frac{C}{n^2}.$$
How to evaluate the series?
I have found a related question.
I apologize for incorrect information in the previous post.
In general $$F(\alpha) = \sum_{n=1}^\infty \frac{\cot(n\pi \alpha)}{n^3}$$ converges and can be explicitly calculated when $\alpha$ is a quadratic irrational. The convergence in this case is easily seen as $\alpha$ has irrationality measure $2$. More precisely, $F(\alpha)/\pi^3 \in \mathbb{Q}(\alpha)$ when $\alpha$ is quadratic irrational.
The procedure below also works when $n^3$ is replaced by any $n^{2k+1}$.
Let $$g(z) = \frac{\cot(z\pi \alpha)\cot(z\pi)}{z^3}$$ then $g$ has simple poles at non-zero integer multiples of $1$ and $1/\alpha$, and $5$-th order pole at $0$. Let $R_N$ denote a large rectangle with corners at $N(\pm 1 \pm i)$. Then contour integration gives $$\tag{1}\sum_{\substack{n\in R_N \\ n\neq 0}} \frac{\cot(n\pi\alpha)}{\pi n^3} + \sum_{\substack{n/\alpha\in R_N \\ n\neq 0}} \frac{\alpha^2\cot(n\pi/\alpha)}{\pi n^3}-\frac{\pi ^2 \left(\alpha^4-5 \alpha^2+1\right)}{45 \alpha} = \frac{1}{2\pi i} \int_{R_N} g(z)dz$$ I claim there exists a sequence of integers $N_1, N_2, \cdots$ such that RHS tends to $0$. Note that $\cot(z\pi)$ is uniformly bounded on the annulus $R_{N+3/4} - R_{N+1/4}$ when $N$ is an integer. Hence by equidistribution of $n\alpha$ modulo $1$, we can find integers $N_i$ such that both $\cot(z\pi\alpha)$ and $\cot(z\pi)$ are uniformly bounded on $R_{N_i+3/4} - R_{N_i+1/4}$.
Since we already know the series converges, from $(1)$: $$\tag{2}F(\alpha) + \alpha^2F(\frac{1}{\alpha}) = \underbrace{\frac{\pi ^3 \left(\alpha^4-5 \alpha^2+1\right)}{90 \alpha}}_{\rho(\alpha)}$$ Note that obviously $F(\alpha+1)=F(\alpha)$.
Let the continued fraction expansion of $\alpha$ be given by $$\alpha = [a_0;a_1,a_2,\cdots]$$ Successive complete quotients are denoted by: $$\zeta_0 = [a_0;a_1,a_2,\cdots]\qquad \zeta_1 = [a_1;a_2,a_3,\cdots]\qquad \zeta_2 = [a_2;a_3,a_4,\cdots]$$ Then $(2)$ and periodicity implies for $k\geq 0$: $$\tag{3} F(\zeta_{k+1}) + \zeta_{k+1}^2 F(\zeta_k) = \rho(\zeta_{k+1})$$ If continued fraction of $\alpha$ is of form $$\alpha = [a_0;a_1,\cdots,a_m,\overline{b_1,\cdots,b_r}]$$ Then $\zeta_{m+r+1} = \zeta_{m+1}$, so we eventually entered a cycle. $(3)$ gives a system of $m+r+1$ linear equations (by setting $k=0,\cdots,m+r$), with $m+r+1$ variables: $F(\zeta_0), F(\zeta_1),\cdots,F(\zeta_{m+r})$. $$\begin{cases} F(\zeta_1) + \zeta_1^2 F(\zeta_0) &= \rho(\zeta_1) \\ F(\zeta_2) + \zeta_2^2 F(\zeta_1) &= \rho(\zeta_2) \\ \cdots \\ F(\zeta_{m+1}) + \zeta_{m+1}^2 F(\zeta_{m+r}) &= \rho(\zeta_{m+1}) \end{cases}$$ Solving it gives the value of $F(\zeta_0)=F(\alpha)$.
For $\alpha = \sqrt{61} = [7;\overline{1,4,3,1,2,2,1,3,4,1,14}]$, we have $$\begin{aligned} \zeta_0 = \sqrt{61} \qquad \zeta_1 &= \frac{1}{12}(7+\sqrt{61}) \\ \zeta_2 = \frac{1}{3}(5+\sqrt{61}) \qquad \zeta_3 &= \frac{1}{4}(7+\sqrt{61})\\ \zeta_4 = \frac{1}{9}(5+\sqrt{61}) \qquad \zeta_5 &= \frac{1}{5}(4+\sqrt{61})\\ \zeta_6 = \frac{1}{5}(6+\sqrt{61}) \qquad \zeta_7 &= \frac{1}{9}(4+\sqrt{61})\\ \zeta_8 = \frac{1}{4}(5+\sqrt{61}) \qquad \zeta_9 &= \frac{1}{3}(7+\sqrt{61}) \\ \zeta_{10} = \frac{1}{12}(5+\sqrt{61}) \qquad \zeta_{11} &= \frac{1}{12}(7+\sqrt{61}) \end{aligned}$$ solving the above system gives the result.
A few examples: for $\alpha = (1+\sqrt{5})/2$, the continued fraction has period $1$, direct substitution into $(2)$ gives $$\sum_{n=1}^\infty \frac{\cot(n\pi \frac{1+\sqrt{5}}{2})}{n^3} = -\frac{\pi ^3}{45 \sqrt{5}}$$ Complexity of result increases as period of $\alpha$ increases. For $\alpha = \sqrt{211}$, which has period $26$: $$\sum_{n=1}^\infty \frac{\cot(n\pi \sqrt{211})}{n^3} = \frac{128833758679 \pi ^3}{383254107060 \sqrt{211}}$$ For $\alpha = \sqrt{1051}$, with period $50$: $$\sum_{n=1}^\infty \frac{\cot(n\pi \sqrt{1051})}{n^3} = \frac{47332791433774124737806821 \pi ^3}{589394448213331173141730140 \sqrt{1051}}$$ When $\alpha$ is not a "pure" quadratic irrational, the result involves "constant term" (because of non-trivial automorphism of $\mathbb{Q}(\alpha)$): $$\sum_{n=1}^\infty \frac{\cot(n\pi(\frac{1}{4}+\frac{\sqrt{7}}{3}))}{n^3} = \frac{13 \pi ^3}{288}+\frac{104771 \pi ^3}{1244160 \sqrt{7}}$$ The closed-form here follows immediately by noting $\csc x = \cot (x/2) - \cot x$.
I wrote a Mathematica code to evaluate this sum. The command
cotsum[Sqrt[61]]evaluates the sum in the question. You can try other quadratic irrationals as well.This algorithm can be made more efficient, but I don't have much motivation to optimize it.