[*Edited to fill in a couple of steps, correct some typos,
standardize the $L$-function notation,
and add comments about the relation with the "Basel sum"*]

Here's a proof via transformation of the sum to
$$
\phantom{(*)CCCCCCCCCCCC}
\left(\frac\pi4\right)^4 + \frac\pi4 L(3,\chi_4) + \frac6\pi L(5,\chi_4),
\phantom{CCCCCCCCCCCC}(*)
$$
where $L(\cdot,\chi_4)$ is the Dirichlet $L$-function defined for $s>0$ by
$$
L(s,\chi_4) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}
= 1 - \frac1{3^s} + \frac1{5^s} - \frac1{7^s} + - \cdots .
$$
Since it is known that $L(3,\chi_4) = \pi^3/32$ and $L(5,\chi_4)=5\pi^5/1536$,
this formula comes to
$$
\left(\frac1{256} + \frac1{128} + \frac5{256}\right) \pi^4 = \frac{\pi^4}{32}
$$
as desired.

It was observed already (implicitly by the proposer, and now as I see
also explicitly in G.Edgar's incipient answer [*later completed*]) that
the numbers $c_2=1$, $c_4=1\times\frac13 + \frac13\times 1$,
$c_6=1\times\frac15 + \frac13\times\frac13 + \frac15\times1$, etc.
are the coefficients of $x^2,x^4,x^6,\ldots$ in the power-series
expansion of $f(x)^2$ where
$$
f(x) = \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} = \frac12 \log\frac{1+x}{1-x}.
$$
Hence the desired sum $c_2^2 + c_4^2 + c_6^2 + \cdots$ is
$(2\pi)^{-1} \int_{-\pi}^\pi |f(e^{it})|^4 dt$ by Parseval's formula.
Now for $0<|t|<\pi$ we have
$$
f(e^{it}) = \frac12 \log\frac{1+e^{it}}{1-e^{it}}
= \frac12 \log \left| \cot \frac{t}{2} \right|
+ \frac{\pi i}{4} \mathop{\rm sgn}t,
$$
whence
$$
\left|f(e^{it})\right|^2 = \frac14 \log^2 \left| \cot \frac{t}{2} \right|
+ \frac{\pi^2}{16}
$$
which is symmetrical under $t \leftrightarrow -t$. Since $\cot(t/2) > 0$
for $0<t<\pi$, we're left with the task of evaluating
$$
\frac1\pi \int_0^{\pi}
\left( \frac14 \log^2 \cot \frac{t}{2} + \frac{\pi^2}{16} \right)^2
dt,
$$
which is to say
$$
\frac1{16\pi} \int_0^{\pi} \log^4 \cot \frac{t}{2} dt
+ \frac\pi{32} \int_0^{\pi} \log^2 \cot \frac{t}{2} dt
+ \frac{\pi^4}{4^4}.
$$
So change variables to $u = \log \cot(t/2)$, which ranges from
$+\infty$ to $-\infty$ as $t$ goes from $0$ to $\pi$. We calculate
$dt = -2 e^u du/(1+e^{2u})$, so for each $m=0,1,2,\ldots$ we have
$$
\int_0^{\pi} \log^{2m} \cot \frac{t}{2} dt
= 2 \int_{-\infty}^\infty u^{2m} \frac{e^u}{1+e^{2u}} du
= 4 \int_0^\infty u^{2m} \frac{e^u}{1+e^{2u}} du,
$$
using in the last step the symmetry of the integrand under
$u \leftrightarrow -u$. Expanding $e^u/(1+e^{2u})$ as
$e^{-u} - e^{-3u} + e^{-5u} - e^{-7u} + - \cdots$,
multiplying each term by $u^{2m}$, and integrating yields
$$
\int_0^{\pi} \log^{2m} \cot \frac{t}{2} dt
= 4 (2m)! \left(
1 - \frac1{3^{2m+1}} + \frac1{5^{2m+1}} - \frac1{7^{2m+1}} + - \cdots
\right)
= 4 (2m)! L(2m+1,\chi_4).
$$
Hence our integral equals (*), **QED**.

*Remarks*: i) The term $(\pi/32) \int_0^{\pi} \log^2 \cot (t/2) dt$
could also have been evaluated by applying Parseval to the Fourier series
of $\log \cot (t/2)$, but I do not see how to get at
$\int_0^{\pi} \log^4 \cot (t/2) dt$ in this way.

ii) The proposer notes that the sum
$c_2^2 + c_4^2 + c_6^2 + \cdots = \pi^4/32$
"looks like some sort of convolution of Euler's sum"
$1 + (1/3^2) + (1/5^2) + (1/7^2) + \cdots = \pi^2/8$,
but the evaluation of $\sum_{m=1}^\infty c_{2m}^2$
does not seem to follow directly from this formula,
even though all the steps in the above derivation were
available to Euler (indeed the values of $L(2m+1,\chi_4)$
are expressed in terms of Euler numbers!). For example,
the same "sort of convolution" applied to
$1 + (1/3^4) + (1/5^4) + (1/7^4) + \cdots = \pi^4/96$
yields a sum that does not (as far as I can see) yield to the same technique,
and might not even be a rational multiple of $\pi^8$.