Show $\tan(n) < n^q$, conjectured $q < 1.1$

This is not a complete answer, but I suppose it might be useful (btw I don't know how to answer this question without using the irrationality measure of $\pi$).

Let $\mu$ be a positive real number such that there are infinitely many rational numbers such that $$ \left\vert\pi - \frac{a}{b}\right\vert < \frac{1}{b^\mu} $$ if $\mu$ is the largest such number then we say that $\mu$ is the irrationality measure of $\pi$. It is known (see https://en.wikipedia.org/wiki/Liouville_number#Irrationality_measure) that $$ 2 \le \mu \le 7.60630853 $$

Let as make the assumption that for the true value $\mu$ of $\pi$ we can find infinitely many $a/b$ such that $a$ is even and $b$ is odd and in addition $\pi > a/b$, (this seems very reasonable but I don't have any idea of how it can be proved) then

$$ 0 < \pi - \frac{2n}{2m+1} < \frac{1}{(2m+1)^\mu} $$

and so $$ \pi/2 + m\pi - \frac{1}{2(2m+1)^{\mu-1}} < n < \pi/2 + m\pi $$

observing that $m \approx n/\pi $ we can finally write $$ n = \pi/2 + m\pi - \frac{\alpha}{n^{\mu-1}} $$ for some bounded $\alpha >0$.

Now we can use the expression for $n$ in the expansion of $\tan x$ about $\frac \pi2$: $$ \tan x = -\frac{1}{x-\pi/2} + \frac{1}{3}\left(x - \frac{\pi}{2}\right) + \frac{1}{45}\left(x - \frac \pi2\right)^3 + \cdots$$ and we get: $$ \tan n = \alpha n^{\mu-1} + O(1) $$

You can see that your conjecture would be true (at least for large enough $n$) if $\mu \le 2.1$ and the assumption above is true.

remarks: If you consider the continued fraction expansion of $\pi$: $$ \pi = 3 + \frac 1{7 + \frac 1{15+\frac 1{1 + \frac 1{292+\dots}}}} $$ Then the candidates to the fractions $a/b$ above are those obtained from stopping the continued fraction at any point, or the same adding 1 to the last partial quotient. If we keep those fractions that verify the assumption above we get the following candidates for $n$:

4                344 0.929204 *
8             260515 1.031030 * 
12           4846147 0.986072 
15         122925461 1.052508 *
17         534483448 1.063489 *
19        3083975227 1.067087 *
22      902209779836 1.026923
26    74357078147863 1.018592 
27   214112296674652 1.087606 *
30 18190586279576483 1.020496
 .....

I have looked for the first 10000 convergents and there isn't any one which gives a larger value than tan(214112296674652). As the number increases it seems to approach 1. For example for the convergents found with 1000 partial quotients or more the largest has exponent 1.0033.

Edit: the code I have used to obtain the above data is the following Pari-GP program:

\\ change to 100 digits precission
\p 100
A = contfrac(Pi);
M=0;  
h=[A[1],1];
k=[1,   0];
{
    for(r=1,length(A)-2,
        u=h;
        h=u*A[r+1]+k;
        k=u;
        if(h[1]%2==0 && Pi> h[1]/h[2], 
            ex=log(tan(h[1]/2))/log(h[1]/2);
            printf("%3d %20d %.8f %d\n", r+1, h[1]/2, log(tan(h[1]/2))/log(h[1]/2),if(ex>M, M=ex;1,0))
        );
        u = u+h;
        if(A[r+2] != 1 && u[1]%2==0 && Pi> u[1]/u[2],
            ex=log(tan(u[1]/2))/log(u[1]/2);
            printf("%3d %20d %.8f %d\n", r+1, u[1]/2, log(tan(u[1]/2))/log(u[1]/2),if(ex>M, M=ex;1, 0))
        )
    );
}