The constants you sought may be expressed in terms of (generalized) hypergeometric functions. As the results are quite involved, I shall only sketch the procedure.
Put
$$\mu_n=\nu_n+\frac6{11}\sum_{\iota=1}^{n-1}\frac1\iota,$$
$\nu_n$ satisfies the homogeneous recurrence. Let
$$f(z)=\sum_{n=1}^\infty \nu_nz^n$$
be the generating function. The recurrence translates to the inhomogeneous DE
$$z^3 (1-z)^3 f'''(z)+z (1-z)^3 f'(z)-(1-2 z+7 z^2) f(z)=z^2 (u+4vz),$$
where
$$\begin{aligned}
&u=\nu_2-\nu_1,&&v=2\nu_3-\nu_2-\nu_1
\end{aligned}$$
depend on initial values. The singularities of this DE lie at $z=0$, $1$, $\infty$. The asymptotic behavior of $\nu_n$ is reflected on the singularity of $f(z)$ at $z=1$. At this point the DE has a holomorphic particular integral $P(z)$, and a set of fundamental solutions of the homogeneous DE are
$$\begin{aligned}
&\frac{z}{1-z},&&F_1(z)=(1-z)^{2+i\sqrt{2}}(1+c_1(1-z)+\cdots),&&F_2(z)=(1-z)^{2-i\sqrt{2}}(1+c_2(1-z)+\cdots).
\end{aligned}$$
If we have
$$f(z)=P(z)+C\frac{z}{1-z}+C_1F_1(z)+C_2F_2(z),$$
then the standard procedure (Darboux's method) yields
$$\nu_n\sim C+C_1\frac{\sinh (\sqrt{2} \pi ) \Gamma (3+i \sqrt{2})}{\pi i}n^{-3-i\sqrt{2}}-C_2\frac{\sinh (\sqrt{2} \pi ) \Gamma (3-i \sqrt{2})}{\pi i}n^{-3+i\sqrt{2}}.$$
In general the coefficients $C$'s can only be determined numerically, as they depend on the special values of these solutions.
In the present case, however, a reduction to the hypergeometric DE is possible. Put
$$f(z)=\frac{z}{1-z}\int g(z)dz,$$
it leads to
$$z^2 (1-z)^2g''(z)+3 z (1-z) g'(z)+(1+4 z+z^2)g(z)=u+4vz.$$
If we further substitute
$$g(z)=z^{-1}(1-z)^{2+i\sqrt{2}}h(z),$$
we find
$$z (1-z) h''(z)+(1-2(1+i \sqrt{2}) z) h'(z)+(1-i \sqrt{2}) h(z)=(1-z)^{-3-i\sqrt{2}}(u+4vz).$$
The left hand side is hypergeometric DE, so we can find a set of fundamental solutions of the homogeneous DE of $g(z)$
$$G_1(z)=\frac{(1-z)^{2+i \sqrt{2}}}{z}\, _2F_1\left(\frac{1}{2}+i \sqrt{2}-\frac{i \sqrt{3}}{2},\frac{1}{2}+i \sqrt{2}+\frac{i \sqrt{3}}{2};1+2 i \sqrt{2};1-z\right),$$
$$G_2(z)=\frac{(1-z)^{2-i \sqrt{2}}}{z}\, _2F_1\left(\frac{1}{2}-i \sqrt{2}-\frac{i \sqrt{3}}{2},\frac{1}{2}-i \sqrt{2}+\frac{i \sqrt{3}}{2};1-2 i \sqrt{2};1-z\right).$$
By variation of constants, we find
$$\begin{aligned}
g(z)&=G_1(z)\int_0^z\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_2(z)dz-G_2(z)\int_0^z\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_1(z)dz\\
&=Q(z)+c_1G_1(z)+c_2G_2(z),
\end{aligned}$$
where
$$Q(z)=-G_1(z)\int_z^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_2(z)dz+G_2(z)\int_z^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_1(z)dz$$
is holomorphic at $z=1$, and
$$\begin{aligned}
&c_1=\int_0^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_2(z)dz,&&c_2=-\int_0^1\frac{iz(u+4vz)}{2\sqrt{2}(1-z)^5}G_1(z)dz.
\end{aligned}$$
Note that these integrals actually diverges, but this difficulty can be circumvented by analytic continuation or by other means.
Finally,
$$f(z)=\frac{z}{1-z}\int_0^zg(z)dz=\frac{z}{1-z}\int_0^1g(z)dz-\frac{z}{1-z}\int_z^1(Q(z)+c_1G_1(z)+c_2G_2(z))dz.$$
It follows that
$$\begin{aligned}
&C=\int_0^1g(z)dz,&&P(z)=-\frac{z}{1-z}\int_z^1Q(z)dz,&&C_iF_i(z)=-c_i\frac{z}{1-z}\int_z^1G_i(z)dz.
\end{aligned}$$
In particular,
$$\begin{aligned}
&C_1=\frac{-3+i \sqrt{2}}{11}c_1,&&C_2=\frac{-3-i \sqrt{2}}{11}c_2.
\end{aligned}$$
In summary, the second constant 0.0565532 can be expressed by integrals involving $\, _2F_1$, or by values of $\, _3F_2$ at $1$. The first constant 0.308492 is expressed by double integrals involving $\, _2F_1$, which I do not know whether it can be reduced further or not.
I retain the dependence on initial values for there is a special case that renders a more exact treatment. If $u=4v=3k$ or $\nu_2=\nu_1+3k$, $\nu_3=\nu_1+15k/8$, the DE of $g(z)$ has a particular solution $Q(z)=k/z$, and the constants in the asymptotic expansion can be expressed by Gamma functions.
Best Answer
Thanks to Peter's comment, I found the solution: $T(n) = n^{\frac{2^k - 1}{2^k}}\cdot T(n^{\frac{1}{2^k}}) + kn$.
Using not 1, but 2 as the stopping point: $n^{\frac{1}{2^k}} = 2$, which means that $k = \log \log n$.
Putting it back to the equation: $T(n) = n^{\frac{\log n - 1}{\log n}}\cdot T(2) + n \log \log n$, which means that the order of growth is $O(n \log \log n)$