As not necessarily proven results were asked for, I have found the following quite accurate:
$$N_k(x):=\ \mid\{n\leq x : \Omega(n)=k\}\mid \ \sim \Re\bigg(\frac{2^{1-k}\alpha e^{1+e}x\log(1+e+\log(2^{1-k}\alpha x))^{\beta}}{\beta!(1+e+\log(2^{1+e}\alpha x)}\bigg) $$ for $1 \leq k\leq \lfloor \log_2 (x) \rfloor$, where $\log_2$ is $\log$ base $2$, $\gamma $ is Euler's constant, $\beta=1+e+ \log \alpha +(1+e+ \log \alpha) ^{1/\pi}$, and$$ \alpha=\frac{1}{2}\ \rm{erfc}\bigg(-\frac{k-(2e^{\gamma}+\frac{1}{4})}{(2e^{\gamma}+\frac{1}{4})\sqrt{2}\ }\bigg)-2\rm{T}\bigg(\bigg(\frac{k}{(2e^{\gamma}+\frac{1}{4})}-1\bigg),e^{\gamma}-\frac{1}{4}\bigg)\\ $$ where $\rm{erfc}$ is the complementary error function and $\rm{T}$ is the Owen T-function.
In integral form, $$\alpha= \frac{1}{\pi}\int_{(-3+8e^\gamma)/(\sqrt{2}(1+8e^\gamma))}^\infty e^{-t^2}\rm{d} t +\int_0^{1/4\ -\ e^\gamma}\frac{e^{-(3\ -\ 8e^\gamma)^2(1+t^2)/(2(1+8e^\gamma)^2)}}{1+t^2}\rm{d} t.$$
As $k\rightarrow \infty$, $\alpha\rightarrow 1$, so
$$\lim_{k \rightarrow \infty}N_{k}(x \cdot 2^{k-1})\sim\frac { {e^{e+1}} x\log\log( {e^{e+1}} x)^{\beta}}{\log( {e^{e+1}} x)\beta!}, $$ where $\beta=\log(e^{e+1})+\log(e^{e+1})^{1/\pi}.$
For $k\leqslant 3$, improvements to the above can certainly be made, but as $k\rightarrow \infty$ (or more correctly, as $k\rightarrow \lfloor \log_2 (x) \rfloor$), the formulae above, as far as have been tested, seem to be fairly accurate.
For convenience, I include the following Mathematica code:
cdf[k_, x_] :=
Re[N[
(2^-k E^(1 + E) x Log[1 + E + Log[2^-k x (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])]]^(1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] +4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])] + (1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])])^(1/\[Pi])) (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma]))/((1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])] + (1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])])^(1/\[Pi]))!
(1 + E + Log[2^-k x (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2] (1 + 8 E^EulerGamma))] +
4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma), 1/4 - E^EulerGamma])]))]];
landau[k_, x_] := N[(x Log[Log[x]]^(-1 + k))/((-1 + k)! Log[x])];
actual[k_, x_] := N[Sum[1, ##] & @@ Transpose[{#, Prepend[Most[#], 1], PrimePi@
Prepend[ Prime[First[#]]^(1 - k) Rest@FoldList[Times, x, Prime@First[#]/Prime@Most[#]],
x^(1/k)]}] &@Table[Unique[], {k}]];
I warmly welcome any criticism or comments on the above, and apologise in advance if I have made any serious errors.