Question. Can this number be expressed in terms of classical values? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^{\frac32}}=1.056348517615643291\dots$$
UPDATE. I'm encouraged by Noam, Kevin and Igor's directional replies. To spice things up, I ask: is this true? $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s),$$ wherever convergence occurs. Here, $\zeta(s)$ and $\beta(s)$ are the Riemann zeta and Dirichlet beta functions, respectively.
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 :-)
This is (almost) rehashing what has been said but with an intent to spell things out.
Begin with the observation that $$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^s}=\frac14\sum_{k=1}^{\infty}\frac{r_2(k)}{k^s}-\zeta(2s).$$ Using $r_2(k)=4\sum_{d\vert k}\left(\frac{-4}k\right)=4(1*\left(\frac{-4}k\right))(k)$, where $\left(\frac{a}b\right)$ is the Jacobi symbol, and $\left(\frac{-4}{2k}\right)=0, \left(\frac{-4}{2k+1}\right)=(-1)^k$, while the Dirichlet series evolves under the arithmetic convolution, i.e., $\sum\frac{(a*b)(k)}{k^s}=\sum\frac{a(k)}{k^s}\sum\frac{b(k)}{k^s}$, it follows that \begin{align}\sum_{k=1}^{\infty}\frac{r_2(k)}{k^s} &=4\sum_{k=1}^{\infty}\frac{(1*\left(\frac{-4}k\right))(k)}{k^s} =4\sum_{n=1}^{\infty}\frac1{n^s}\sum_{k=1}^{\infty}\frac{\left(\frac{-4}k\right)(k)}{k^s} \\ &=4\zeta(s)\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^s}=4\zeta(s)\beta(s). \end{align}
Exhaustive analysis of lattice sums can be found in the book Lattice sums then and now. This book can be found for free on the web.
OP's sum is given by formula 1.3.14 on page 33:
L. Lorenz. Bidrag tiltalienes theori. Tidsskrift Math., 1:97–114, 1871.
To give yet a third version of Kevin and Noam's answer, check out Paul Garett's notes where he shows how to factorize the Zeta function of the Gaussian integers.
