By numerical experimentation I find the first three terms of the Puiseux series of the Bessel function of the first kind
$$
J_n(n) =
\frac{\Gamma(\frac13)}{2^{2/3}\cdot 3^{1/6} \cdot \pi}n^{-1/3}
– \frac{1}{35\cdot 6^{1/3}\cdot\Gamma(\frac13)}n^{-5/3}
– \frac{\Gamma(\frac13)}{225 \cdot 2^{2/3}\cdot 3^{1/6}\cdot \pi}n^{-7/3}
+\mathcal{O}(n^{-11/3})
$$
How does this series continue?
See also this application.
How I got this far
first term
For the first term, start with the integral representation
$$
J_n(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi} d\theta \cos[n(\sin(\theta)-\theta)]
$$
For $n\to\infty$ the only significant contributions to this integral come from values of $\theta$ that are close to zero. Therefore we approximate $\sin(\theta)-\theta\approx-\theta^3/6$ and find
$$
\lim_{n\to\infty} n^{1/3}\cdot\frac{1}{2\pi}\int_{-\pi}^{\pi} d\theta \cos[-n\theta^3/6] = \frac{\Gamma(\frac13)}{2^{2/3}\cdot3^{1/6}\cdot\pi}
$$
In Mathematica:
Limit[1/(2π) Integrate[Cos[n (-(θ^3/6))], {θ, -π, π}]*n^(1/3), n -> ∞]
Gamma[1/3]/(2^(2/3) 3^(1/6) π)
second term
In Mathematica, define the Bessel function and its one-term approximation, as well as their numerical difference evaluated to 1000 digits:
b[n_] = BesselJ[n, n];
ba[n_] = Gamma[1/3]/(2^(2/3)*3^(1/6)*π)*n^(-1/3);
B[n_] := N[b[n] - ba[n], 10^3]
Calculate how the numerical difference behaves for large $n$ (after multiplying it by $n^{5/3}$):
ListLinePlot[T = Table[B[n]*n^(5/3), {n, 10^Range[2, 5, 1/4]}]]
and find the approximate numerical value of the limit as $n\to\infty$:
NumericalMath`NSequenceLimit[T]
-0.00586928848357833870
Then use AskConstants to find that this number is probably equal to $-\frac{1}{35\cdot 6^{1/3}\cdot\Gamma(\frac13)}$.
third term
Same procedure as second term, but with the better approximation
ba[n_] = Gamma[1/3]/(2^(2/3)*3^(1/6)*π)*n^(-1/3) -
1/(35*6^(1/3)*Gamma[1/3])*n^(-5/3);
and multiplying the difference B[n]
by $n^{7/3}$ before taking the numerical limit $n\to\infty$. The result is $-0.0019880325262065435671$, which AskConstants thinks is equal to $- \frac{\Gamma(\frac13)}{225 \cdot 2^{2/3}\cdot 3^{1/6}\cdot \pi}$.
higher-order terms
The above recipe can be continued to higher-order terms, but I lose confidence in the capabilities of AskConstants. The fourth term is $+0.00048679979012516409164$, which may be
$$
+\frac{1213}{511875\cdot6^{1/3}\cdot \Gamma(\frac13)}n^{-11/3}
$$
but such large rationals don't inspire confidence.
Best Answer
The answer is given by Ernst Meissel in this 1891 paper (in German):
$$ J_n(n) = \frac{1}{\pi} \sum_{m=0}^{\infty} \lambda_m \cdot \Gamma\left(\frac{2m+4}{3}\right) \cdot \left(\frac{6}{n}\right)^{\frac{2m+1}{3}} \cdot \cos\left(\frac{2m+1}{6}\pi\right) $$
The coefficients $\lambda_m$ describe the Taylor series of the solution $u(x)=\sum_{m=0}^{\infty} \lambda_m x^{2m+1}$ of the transcendental equation $u-\sin(u)=x^3/6$ around $x=0$. Term-by-term comparison gives
$$ \lambda_0=1\\ \lambda_1=\frac{1}{60}\\ \lambda_2=\frac{1}{1400}\\ \lambda_3=\frac{1}{25200}\\ \lambda_4=\frac{43}{17248000}\\ \lambda_5=\frac{1213}{7207200000}\\ \lambda_6=\frac{151439}{12713500800000}\\ \lambda_7=\frac{33227}{38118080000000}\\ \lambda_8=\frac{16542537833}{252957982717440000000}\\ \lambda_9=\frac{887278009}{177399104762880000000}\\ \lambda_{10}=\frac{15233801224559}{39217856135377920000000000}\\ \ldots $$
These coefficients can be calculated efficiently with the Mathematica code
Or all at once by series inversion (thanks to J.M.): calculate $\lambda_0\ldots\lambda_n$ with
Thanks to J.M. who pointed out Meissel's paper to me.