We know that Skewes' number is a yet unknown huge integer $x$ such that,
$$\pi(x) > \operatorname{li}(x)\tag1$$
where $\pi(x)$ is the prime counting function and $\operatorname{li}(x)$ is the logarithmic integral function. We also know that the $\text{RHS}$ is just the first term of an infinite series as stated by Ramanujan and Riemann,
$$\pi(x) = \operatorname{li}(x) -\tfrac12\operatorname{li}(x^{1/2})-\tfrac13\operatorname{li}(x^{1/3})-\tfrac15\operatorname{li}(x^{1/5}) +\dots$$
Q: What if we also use and modify the second term, $$\pi(x_n) > \operatorname{li}(x_n) - \tfrac1{\color{blue}n} \operatorname{li}(x_n^{1/2})\tag2$$ and for a given $n$, find the smallest integer $x>8$ that satisfies the inequality?
For $n=2,3,4,5$, we have,
$$x_2 = 19$$ $$x_3 = 59753$$ $$x_4 = 30902129$$ $$x_5 = 110087953$$
with $2,5,8,9$ digits and $x_5$ found by DanaJ. Of course, as $n$ climbs ever higher and $x_n$ reaches $317$ digits, it will then reach Skewes' number which is true for all $n$.
So what is $x_6$ and higher?
I used the following:
where you can change the '5' to whatever is desired. It returns the OP's values for 2-5. About 11 seconds on idle Macbook, single thread for n=5, very roughly 12 hours for n=6. Bypassing Perl we can do this 2-4x faster.
We're walking primes and checking li each time. primesieve is a bit faster at sieving, but it's a very small portion of the time. I also wrote a ~ 15 line C version that runs 2-4x faster as it avoids the overhead of going back and forth between Perl and C. A faster version of li() would be the most obvious improvement. As it is, finding $x_5$ takes about 3.3 seconds, and $x_6$ takes 2.5 hours. These could be parallelized.
We can speed it up a lot using some bounds on li growth to avoid calling the li() function much of the time. This yields 4.5 minutes for $x_6$. Calling li() is most of the time vs sieving but we're getting closer. Sieving is all very fine for smallish inputs, but at some point it's going to take appreciable time (1-2 hours to 10^13, about 12x longer per 10x farther sieving).
There's a very important issue as x gets large: the accuracy of li(). While the error for the code I'm using is under 5e-17 relative error, with large enough x this is significant. Enough to make the $x_7$ bound harder to determine. We can use mpf or quad floats but this has a serious time and complexity hit. I have quadmath (GNU true 128-bit floats) and mpf/mpfr code for li() -- the former has absolute error ~ 0.001 at 1e19, so much better for $x_7$ than double or Intel long double.
In the table, let, $$K_n=\text{Kulsha's maxima}$$ $$\text{dif}_n=-\pi(x_n) + \operatorname{li}(x_n) - \tfrac1{n} \operatorname{li}(x_n^{1/2})$$
Skewes' number will yield negative $\text{dif}_n$ for all $n$.
$$\begin{array}{|c|l|l|c|c|c|} \hline n&x_n& K_n& \approx (x_n)&\approx (K_n-x_n) &\text{dif}_n\\ \hline 2&19& 19&1.9\times10 &0&-0.041\\ 3&59753&-&5.9\times10^4&-&-0.081\\ 4&30902129& 30909673&3.09\times10^7&7\times10^3&-0.072\\ 5&110087953& 110102617&1.10\times10^8&1\times10^4&-0.192\\ 6&330946474073& 330957852107&3.309\times10^{11}&1\times10^7&-0.235\\ 7&1324783721133846733\,\color{red}?& 1325005986250807813&1.32\times10^{18}&2\times10^{14}&-0.379\\ \hline \end{array}$$
I also note that the values for 2, 4, 5, and 6 are quite close to the local maximas shown on Kulsha's page. For $x_7$, I didn't find anything around the 10^16 to 10^17 maxima (m-1e11,m+2e9), but an upper bound is found near the 10^18 maxima. There is no smaller $x_7$ within $10^{13}$ of the found result.
More details in Tito's comments.