I don't know how "classical" you find these values, but here's perhaps something.
Define $E=\sum_{m,n\in\mathbb{Z}}q^{m^2+n^2}$, which is known to be a weight 1 level 4 modular form. In fact $E$ is an eigenform for the Hecke operators, and if we write $E=\sum_{r\geq0}a_rq^r$ then $L(E,s)=\sum_{r\geq1}a_r/r^s$ equals $4\zeta(s)L(\chi,s)$ with $\chi$ the Dirichlet character of conductor 4. The factor of 4 is because $E$ is normalised so that $a_1=4$. Note that $a_r$ equals the number of ways $r$ can be written as $m^2+n^2$ (where we allow zero and negative values for $m$ and $n$ at this point). Note also that the $L$-function doesn't see the troublesome $m=n=0$ term.
Now note that setting $s=3/2$ (where everything converges) we get $L(E,3/2)$ is nearly what you want. In fact if your constant is $c$ then (allowing for signs) we get
$$4c+4\sum_{n\geq1}(1/n^2)^{3/2}=L(E,3/2)$$
($4c$ for the signs, and the other term for the $m=0$ and $n=0$ terms we missed out) and all the 4s cancel miraculously giving
$c=\zeta(3/2)L(\chi,3/2)-\zeta(3)$
or equivalently
$c=\zeta_{\mathbb{Q}(i)}(3/2)-\zeta(3)$
with $\zeta_{\mathbb{Q}(i)}$ the Dedekind zeta function of the number field. Let's check with pari-gp:
L=lfuncreate(x^2+1);
lfun(L,3/2)-zeta(3)
%2 = 1.0563485176156432910328906583178146441
which looks good to me.
Note finally that now I've done the calculation I realise that one could avoid the modular forms side of things and just consider the zeta function of $\mathbb{Q}(i)$ directly because we're summing some function of norms of elements; the final equation for $c$ follows essentially from the definition of the Dedekind $\zeta$ function.
Hmm, and now note finally finally that while I was typing this, Noam Elkies said exactly the same thing but rather more succinctly :-)