The formula is
$$d = 6\,n - \sum_{i= 1}^{J}\left( 6 - f_i\right)$$
or
$$d = 6N - 1 - \sum_{i=1}^j\ (6 - f_i) = 6(N-1 - J) + \sum_{i=1}^J\ f_i $$
In the book Modern Robotics, there is an additional (-1) in the formula, as shown in the second version of the formula, as they recommend counting the ground as linkage. I do not count ground as linkage, I do not use the -1. For me it is just more logical this way, since ground is not a linkage and then I do not need to subtract 1 form the number of linkages. Furthermore the constant 6 in this version of the formula is denoted by m in the book Modern Robotics.
let's denote the number of chains grasping the object with $c$ in order to avoid confusion with the already used $n$ notation.
$N$ is the number of links including the ground link,
$n$ is the number of links not including the ground link,
$J$ is the number of joints and
$fi$ is the number of degrees of freedom of joint $i$.
so fort of all we have 1 linkage, the object being held commonly (assuming the grasp is rigid) counts as a linkage.
Per kinematic chain we have 3 linkages, we have to subtract 1 since the last one is grasping, therefore is part of the first linkage. In other words adding another kinematik chain increases the number of linkages by 2 and not by three, since the last linkage will be part of the rigid body, there is no motion between them, therefore it does not count as a new linkage.
$n = 1 + 2 \cdot c$ for the case in the picture, for $c = 4$ linkages $n = 9$
There are 3 joints per kinematic chain. one of the joints with 3 dof, one of the joints with 1 dof and the last joint ... is tricky, it seems to have 3 dof, but in only counts as 2, since the two spherical joints produce a local rotation motion around the axis between their middle points. This a degree of freedom, which does not effect the motion of the whole chain, therefor it needs to be subtracted (if the number of kinematic chains is greater then 1, $c > 1$). So:
$$\sum_{i= 1}^{J}\left( 6 - f_i\right) = c \cdot ((6 - 3) + (6 - 1) + (6 - 2)) = c \cdot (3 + 5 + 4) = c \cdot 12 = 4 \cdot 12 = 48$$
Therfore, for the mechanism in the picture:
$d = 6 \cdot 9 - 48 = 54 - 48 = 6$
For the generalized version, for $c > 1$ number of kinematic chains added:
$d = 6 \cdot (1 + 2 \cdot c) - (c \cdot 12) = 6 + 12 \cdot c - 12 \cdot c$
$d = 6$, independent of the number of chains used.
It makes scene, since one chain has 6 degrees of freedom, if you couple the same chain parallel, is should not loose of gain any degrees of freedom.
If we would add 6 linkages, then we would obtain a structure very similar to the Stewart platform. It is commonly known that the Stewart platform has 6 DOF.
