A similar approach to my previous solution above but neater:
Using the well-known identity
$$\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$$
Divide both sides by $x$ then integrate , we get
$$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+C $$
set $x=0,\ $ we get $C=2\ln2$
Then
$$\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+2\ln2\tag1$$
Multiply both sides of (1) by $-\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $1$ and use the fact that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
\begin{align}
\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}&=2\underbrace{\int_0^1\frac{\ln(1+\sqrt{1-x})\ln(1-x)}{x}dx}_{\sqrt{1-x}=y}-2\ln2\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{-\zeta(2)}\\
&=8\int_0^1\frac{y\ln(1+y)\ln y}{1-y^2}dy+2\ln2\zeta(2)\\
&=4\int_0^1\frac{\ln(1+y)\ln y}{1-y}-4\int_0^1\frac{\ln(1+y)\ln y}{1+y}+2\ln2\zeta(2)
\end{align}
where the first integral is
$$\int_0^1\frac{\ln y\ln(1+y)}{1-y}\ dy=\zeta(3)-\frac32\ln2\zeta(2)$$
and the second integral is
$$\int_0^1\frac{\ln y\ln(1+y)}{1+y}\ dy=-\frac12\int_0^1\frac{\ln^2(1+y)}{y}dy=-\frac18\zeta(3)$$
Combine the results of the two integrals we get
$$\boxed{\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)}$$
If we differentiate both sides of $\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
$$ \int_0^1x^{n-1}\ln x\ln(1-x)dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}-\zeta(2)}{n}\tag2$$
Now multiply both sides of $(2)$ by $ \frac{1}{4^n}{2n\choose n}$ the sum up from $n=1$ to $\infty$ we get
$$\sum_{n=1}^\infty \frac{H_n}{n^24^n}{2n\choose n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}-\zeta(2)\sum_{n=1}^\infty \frac{1}{n4^n}{2n\choose n}\\=\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\=\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx}_{\text{Beta function:}7\zeta(3)-6\ln2\zeta(2)}-\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x}dx}_{\zeta(3)}$$
Substitute $\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)$ and $\sum_{n=1}^\infty\frac{1}{n4^n}{2n\choose n}=2\ln2$ we get
$$\boxed{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n4^n}{2n\choose n}=\frac32\zeta(3)}$$
Using the identity
$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}\tag3$$
Again multiply both sides of $(3)$ by $ \frac{1}{4^n}{2n\choose n}$ the sum up from $n=1$ to $\infty$ we get
$$\sum_{n=1}^\infty \frac{H_n^{2}}{n4^n}{2n\choose n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}\\=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx=\int_0^1\frac{\ln^2(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\=\underbrace{\int_0^1\frac{\ln^2(1-x)}{x\sqrt{1-x}}dx}_{\text{Beta function:}14\zeta(3)}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}dx}_{2\zeta(3)}$$
Finally, substitute $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}=\frac32\zeta(3)$ we get
$$\boxed{\sum_{n=1}^\infty \frac{H_n^{2}}{n4^n}{2n\choose n}=\frac{21}2\zeta(3)}$$
Addendum:
Above, We calculated two integrals using Beta function but we can do them in a different way:
For the first integral $$\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx=\int_0^1\frac{\ln(1-x)\ln x}{(1-x)\sqrt{x}}dx=\int_0^1\frac{x^{-1/2}\ln x\ln(1-x)}{1-x}dx$$
we can use the generalization
$$\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right)$$
set $m=1$ then let $n$ approach $-1/2$ we get
$$\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx=7\zeta(3)-6\ln2 \zeta(2)$$
For the second integral, set $\sqrt{1-x}=y$
$$\int_0^1\frac{\ln^2(1-x)}{x\sqrt{1-x}}dx=8\int_0^1\frac{\ln^2y}{1-y^2}dy\\=8\sum_{n=0}^\infty\int_0^1 x^{2n}\ln^2y\ dy=16\sum_{n=0}^\infty\frac{1}{(2n+1)^3}=16\left(\frac{7}{8}\zeta(3)\right)=14\zeta(3)$$
A Little bonus:
Multiply both sides of $\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$ by $-\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $1$ and use the fact that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get
$$\sum_{n=1}^\infty\frac{H_n}{n4^n}{2n\choose n}=-\underbrace{\int_0^1\frac{\ln(1-x)}{x\sqrt{1-x}}dx}_{\sqrt{1-x}=y}+\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{-\zeta(2)}\\=-4\int_0^1\frac{\ln y}{1-y^2}dy-\zeta(2)=-4\left(-\frac34\zeta(2)\right)-\zeta(2)=\boxed{2\zeta(2)}$$