Some hefty circuit!

First find the equivalent resistance between $A$ and $B$ . By symmetry, all nodes on the middle diagonal must have the same potential, so we can connect them all without changing voltages/currents. The resulting network consists of two identical triangle networks in series, so we only need to consider the top one.

Again by symmetry, currents on the left of the triangle must equal currents on the right of the triangle: we may split the triangle in half and have two identical networks in parallel. Again, we only need to consider one of them. This is what we are left with:

We calculate the equivalent resistance between nodes A and M via nodal analysis to get $R_{AM} = 1 + R_{1M}$ (see program below). Using all symmetry, we get $R_{AB} = 2 (R_{AM}\|R_{AM}) = R_{AM} = 1 + R_{1M}= \frac{6385}{2494}\qquad\Rightarrow\qquad I = \frac{V}{R_{AM}} = \frac{5051597}{9871210}\approx\boxed{0.512}$

---

/* maxima program to calculate node 
voltages of simplified network  via nodal analysis */

Y : matrix(
    [3/2,  -1,   0,   0, -1/2,  0,  0,    0],
    [ -1,   3,  -1,   0,   -1,  0,  0,    0],
    [  0,   -1,  3,  -1,    0, -1,  0,    0],
    [  0,   0,  -1, 8/3,    0,  0, -1,    0],
    [-1/2, -1,   0,   0,    3, -1,  0, -1/2],
    [  0,   0,  -1,   0,   -1,  4, -1,   -1],
    [  0,   0,   0,  -1,    0, -1,  4,    0],
    [  0,   0,   0,   0, -1/2, -1,  0,  7/2]
)$

J : transpose([1, 0, 0, 0, 0, 0, 0, 0])$

/* equivalent resistance between A and M */
R_AM : 1 + (invert(Y) . J)[1, 1];

The effective resistance between A and B is $\frac{6385}{2494}$ , so $I = \frac{4051/3092}{6385/2494} = \frac{5051597}{9871210} \approx 0.51175.$ Reference: http://personal.rhul.ac.uk/uhah/101/ResPap_08.pdf .