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.
cotsum[x_] /; QuadraticIrrationalQ[x] :=
Module[{a1 = x, list, l, r, i, nlist, solution, output, equation,
string}, list = ContinuedFraction[a1];
l = Length[list] - 1;
r = Length[list[[l + 1]]]; Global`f[a_] := (1 - 5 a^2 + a^4)/90/a;
i = 1; string = "{";
While[i < l + r + 1, string = string <> "x" <> ToString[i] <> ",";
i++]; string = StringTake[string, StringLength[string] - 1] <> "}";
Do[Evaluate[ToExpression["a" <> ToString[i + 1]]] =
FromContinuedFraction[Drop[list, i]], {i, 1, l - 1}];
nlist = list[[l + 1]];
Do[Evaluate[ToExpression["a" <> ToString[i + l + 1]]] =
FromContinuedFraction[{Flatten[
Append[Drop[nlist, i], Take[nlist, i]]]}], {i, 0, r - 1}];
equation =
Table[ToExpression[
"x" <> ToString[i + 1] <> "+a" <> ToString[i + 1] <> "^2*x" <>
ToString[i] <> "==f[a" <> ToString[i + 1] <> "]"], {i, 1,
r + l - 1}];
equation =
Append[equation,
ToExpression[
"x" <> ToString[l + 1] <> "+a" <> ToString[l + 1] <> "^2*x" <>
ToString[r + l] <> "==f[a" <> ToString[l + 1] <> "]"]];
solution = Solve[equation, ToExpression[string]]; Clear["a*"];
output = (ToExpression[string][[1]] /. Flatten[solution])*Pi^3;
Clear[f];
FullSimplify[output]]
