Prove that $A$ is bounded operator on $\ell^p$ and find $\| A\|$

As mention by robjohn, only the case $p>1$ allows $A$ to be bounded.

In your last estimate, you changed $m$ to $\infty$, and there was no reason for that step, as you already had an upper bound.

One way to get a lower bound for the norm of $A$ is as follows. Fix $k\in\mathbb N$. Let $$ x_n=n^{-q/2p}. $$ Then, since $-q/2p-1/2=-q/2$, \begin{align} \|Ax\|_p^p&=\sum_{m=1}^\infty|(Ax)_m|^p =\sum_{m=1}^\infty \left|\frac1m\,\sum_{n=1}^m\frac{n^{-q/2p}}{n^{1/2} }\right|^p\\ \ \\ &=\sum_{m=1}^\infty \left(\frac1m\,\sum_{n=1}^m{n^{-q/2}}\right)^p\\ \ \\ &=\sum_{m=1}^\infty \frac1{m^p}\left[\left(\sum_{n=1}^m{|n^{-q/2p}|^p}\right)^{1/p}\left(\sum_{n=1}^m{n^{-q/2}}\right)^{1/q}\right]^p\\ \ \\ &=\|x\|_p^p\,\sum_{m=1}^\infty\frac1{m^p}\left(\sum_{n=1}^m{n^{-q/2}}\right)^{p/q}. \end{align} So $$ \|A\|\geq \left(\sum_{m=1}^\infty\frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}\right)^{1/p}, $$ as you already had the upper bound.

The series converges for any $p>1$. Indeed, for $m$ large enough we have $$ \sum_{n=1}^mn^{-q/2}\leq 1+\int_1^mt^{-q/2}=1+\frac{m^{1-q/2}-1}{1-q/2}\leq c\,m^{1/2} $$ for some $c>0$. So $$ \frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}\leq c^{1/q}\,m^{p/2q-p} =c^{1/q}\,m^{-p/2-1/2}. $$ Since $p>1$, the series $\sum_{m=1}^\infty\frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}$ converges.

Brute force and Hölder give us, with $\frac1p+\frac1q=1$, $$ \begin{align} \sum_{m=1}^\infty\left|\frac1m\sum_{n=1}^m\frac{x_n}{\sqrt{n}}\right|^p &\le\sum_{m=1}^\infty\frac1{m^p}\left(\sum_{n=1}^m\frac{|x_n|}{\sqrt{n}}\right)^p\\ &\le\sum_{m=1}^\infty\frac1{m^p}\|x_n\|_p^p\left(\sum_{n=1}^m\frac1{n^{q/2}}\right)^{p/q}\\ &\le\|x_n\|_p^p\sum_{m=1}^\infty\frac1{m^p}\left(2\sqrt{m}\right)^{p-1}\tag{$q\ge1$}\\ &=2^{p-1}\|x_n\|_p^p\zeta\left(\frac{p+1}{2}\right) \end{align} $$ Thus, $A$ is bounded for $p\gt1$ with norm no greater than $2^{1-1/p}\zeta\left(\frac{p+1}{2}\right)^{1/p}\le2\,\zeta\left(\frac{p+1}{2}\right)$.

For $(x_n)=(1,0,0,\dots)\in\ell^1$, $$ \frac1m\sum_{n=1}^m\frac{x_n}{\sqrt{n}}=\frac1m $$ Thus, we cannot have that $A$ is bounded on $\ell^1$.