ORIGINAL ANSWER
To put it succinctly.
\begin{align*}
\lim_{N\rightarrow \infty} \sum_{n=1}^{N-1}{\zeta \left( 2\left( N-n \right) \right) \frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}
=&\lim_{N\rightarrow \infty} \sum_{n=1}^{N-1}{\sum_{k=1}^{\infty}{\frac{k^{2n}}{k^{2N}}}\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}
\\
=&-\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}}=-\sin \left( 1 \right)
\end{align*}
If you found this a bit too handwavy, I could provide the solution in greater detail.
EDIT
I shall provide a more rigorous proof here.
To start with, we rewrite the sum in terms of the definition of riemann zeta function. So we have
$$
\sum_{n=1}^{N-1}{\zeta \left( 2\left( N-n \right) \right) \frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}=\sum_{n=1}^{N-1}{\sum_{k=1}^{\infty}{\frac{k^{2n}}{k^{2N}}}\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}
$$
We can interchange the sums if each terms in the finite sum converges, which is obvious in this case. Therefore
$$
\sum_{n=1}^{N-1}{\sum_{k=1}^{\infty}{\frac{k^{2n}}{k^{2N}}}\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}=\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}
$$
Now we have to proof that
$$
\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}=0
$$
Since we have
$$
\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}=\sin \left( k \right)
$$
Which converges, by something Cauchy (Cannot remember the exact name)
$\forall \varepsilon >0, \exists N\in \mathbb{N}\,\,\mathrm{s}.\mathrm{t}.$
\begin{align*}
&\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}<\frac{\varepsilon}{\zeta \left( 2 \right)}<\frac{\varepsilon}{\zeta \left( 2N-1 \right)}
\\
\Rightarrow &\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}<\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\frac{\varepsilon}{\zeta \left( 2N-1 \right)}=\varepsilon
\end{align*}
Which proves the claim. Hence
\begin{align*}
&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}+0
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}+\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}
\\
=&\lim_{N\rightarrow \infty} \left( \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}+\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} \right)
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}
\\
=&-\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}}
\end{align*}
Now, notice that
$$
\sum_{k=1}^{\infty}{\left| \frac{\sin \left( k \right)}{k^{2N-1}} \right|}\leqslant \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}=\zeta \left( 2N-1 \right)
$$
Meaning the sum converge absolutely, implying the sum converges. So we only need to prove
$$
\lim_{N\rightarrow \infty} \left| \sum_{k=2}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|=\lim_{N\rightarrow \infty} \left| \sin \left( 1 \right) -\sum_{k=1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|=0
$$
The gist to prove this is along these lines
\begin{align*}
&\forall \varepsilon >0, \exists N\in \mathbb{N} \,\, \mathrm{s}.\mathrm{t}.
\\
&\left| \sum_{k=2}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|\leqslant \left| \sum_{k=2}^N{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|+\left| \sum_{k=N+1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|\leqslant \sum_{k=2}^N{\frac{1}{k^N}}+\sum_{k=N+1}^{\infty}{\frac{1}{k^2}}<\sum_{k=2}^N{\frac{\varepsilon}{2^{k-1}}}+\frac{\varepsilon}{2^N}=\varepsilon
\end{align*}
By utilizing that cauchy stuff (name forgotten) and the basic definition of limits.
Then, we are done.
A fast and simple solution idea by Cornel Ioan Valean
We will use the power of the ideas and strategies from the books (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023).
Okay, let's start!
From the book, More (Almost) Impossible Integrals, Sums, and Series (2023), Sect. $4.5$, pages $396$-$398$, we have that
$$\displaystyle \color{blue}{-\frac{\log(1-x^2)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}x^{2n} \frac{1}{4^n}\binom{2n}{n}(2 H_{2n}-H_n),\ |x|<1} ,$$
and exploiting this fact and turning the left-hand side into a double integral, $\displaystyle \int _0^{\pi/2}\left(\int _0^{\sin (x)}\frac{\log \left(1-y^2\right)}{y\sqrt{1-y^2}}\textrm{d}y\right)\textrm{d}x$, immediately reveals (after changing the integration order) that
$$\sum _{n=1}^{\infty } \binom{2 n}{n}^2 \frac{2 H_{2 n}-H_n}{n 2^{4 n}}$$
$$ =\frac{\pi^2}{2}+\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1-x)}{x} \textrm{d}x+\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1+x)}{x} \textrm{d}x$$
$$-\frac{16}{\pi }\int_0^1 \frac{\arctan(x) \log(1+x^2)}{x} \textrm{d}x. \tag1$$
On the other hand, exploiting that
$$\int_0^{\pi/2} \log(\sin(x)) \sin^{2n}(x) \textrm{d}x=\frac{\pi}{2}\frac{1}{2^{2n}}\binom{2n}{n}\left(H_{2n}-H_n-\log(2)\right),$$
which is also found in the sequel, page $191$, multiplying both sides by $\displaystyle \frac{1}{n 2^{2n}}\binom{2n}{n}$, making the summation from $n=1$ to $\infty$, and rearranging, we get that
$$\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{H_{2n}-H_n-\log(2)}{n 2^{4n}}$$
$$\small =-\frac{8 }{\pi }\int_0^1 \frac{\log ^2(1+x^2)}{1+x^2} \textrm{d}x+\frac{8 }{\pi } \log (2)\int_0^1 \frac{ \log(1+x^2)}{1+x^2}\textrm{d}x$$
$$+\frac{8 }{\pi }\int_0^1 \frac{\log(x) \log(1+x^2)}{1+x^2} \textrm{d}x.\tag2$$
The rest is known and trivial, and the separate series involving the numbers $H_{2n}$ and $H_n$ are extracted by a system of relation consisting of $(1)$ and $(2)$.
End of story
Thank you so much Cornel for your work and your life-changing books, (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023).
(Very) important: Exploiting such techniques involving the series in blue for building systems of relations, one can derive extremely difficult series. Some (simple) paper(s) showing such examples will be written soon.
Update 1: Indeed, the series taken apart also contain a Trilogarithm with a complex argument. The forms of the two key identities (that is, using integrals) avoid that appearance, and to get directly the desired value without touching the Trilogarithm with a complex argument, all we need is to turn the integrals from the identity in $(2)$ into arctan-log integrals. Then we need to exploit the arctan-log integral identities given in the sequel, Sects. $1.36$-$1.38$, pages $48$-$51$. That's all (and done)!
Update 2: A full solution with all the details of this problem will be found soon in a separate paper (the main series plus the two separated series).
Update 3: Binoharmonic Series with the Squared Central Binomial Coefficient And Their Integral Transformation Using Elliptic Integrals by Cornel Ioan Valean
Best Answer
Given the OP's formula A for $\zeta(2)$, there is a very similar one B for $\zeta(4)$,
\begin{align} A &=\sum_{n=1}^{\infty}\binom{4n-2}{2n-1}\binom{2n}{n}^{-4}\frac{(60n^{2}-43n+2^3)}{n^{4}}\\ A &=\frac{\pi^2}3=2\,\zeta(2)\\[8pt] B &=\sum_{n=1}^{\infty}\binom{4n-3}{2n-1}\binom{2n}{n}^{-8}\frac{(5460n^4-8341n^3+2^8(19)n^2-2^8(5)n+2^7)}{n^8}\\ B &=\frac{\pi^4}{30}=3\,\zeta(4) \end{align}
derived by Almkvist by
flippingshifting the sole Ramanujan-type formula for $1/\pi^4$ found by Jim Cullen in 2010. (The terminology "shifting" is used by Guillera in "Ramanujan Series with a Shift".)I wonder if, using the OP's formula for $\pi^2$, it may be possible to shift it to get a Ramanujan-type formula for $1/\pi^2$?
I. Update (an hour later)
Turns out Formula A is indeed a shifted version of one Guillera's formulas for $1/\pi^2$. Given the polynomial in $A$,
$$60(n+k)^2-43(n+k)+8$$
Let $k=1/2$, and it transforms to $\frac12(120n^2+34n+3)$. We do the same to the polynomial in $B$, and we get two Ramanujan-type formulas,
\begin{align} \frac1{\pi^2} &=\frac{1}{2^{5}}\,\sum_{n=0}^\infty b_3(n)\,\frac{120n^2+34n+3}{2^{4n}}\\[4pt] \frac1{\pi^4} &=\frac{1}{2^{11}}\sum_{n=0}^\infty c_2(n)\frac{43680n^4+20632n^3+4340n^2+466n+21}{2^{12n}} \end{align}
where,
$$b_3(n)=\frac{\big(\tfrac12\big)_n^3\,\big(\tfrac14\big)_n\,\big(\tfrac34\big)_n}{n!^5}, \quad c_2(n)=\frac{\big(\tfrac12\big)_n^7\,\big(\tfrac14\big)_n\,\big(\tfrac34\big)_n}{n!^9} \quad $$
with Pochhammer symbol $(x)_n$.
To recap, the formula for $1/\pi^4$ was found by Cullen and Almkvist
flippedshifted it to $\pi^4$. For this post, it was the reverse: the OP found $\pi^2$ and, if we shift it, we recover one of Guillera's formulas for $1/\pi^2$. Suggested further reading for two of Guillera's papers can be found in the comments.II. Addendum (days later)
"Shifting" an identity does not necessarily lead to a shifted value that is the reciprocal of the first. It may be a totally different value. For example, we use the one by Setness Ramesory in the comments and express it with Pochhammer symbols $(x)_n$,
$$F(k)=\small\sum_{n=1}\frac{\big(\tfrac12+k\big)_n\,\big(1+k\big)_n^3}{\big(\tfrac13+k\big)_n^2\,\big(\tfrac23+k\big)_n^2}\, \frac{145(n+k)^2-104(n+k)+18}{(n+k)^3\,(2(n+k)-1)}\, \left(\frac{2}{27}\right)^{2(n+k)}$$
If $k=0$, then naturally we get Ramesory's,
$$F(0) = \frac{\pi^2}{3}\qquad$$
But if $k=\frac12$, then the shifted value is not the reciprocal, but $\pi$ is an exponent lower and with an added rational,
$$\qquad F\big(\tfrac12\big) = \frac{4\pi}{3}-14\left(\frac23\right)^3$$