Order in the court ....

Same as of Brian Charlesworth

There probably is a more elegant way to do this, by for what it's worth, here is my case by case approach.

(i) If $(a,b) = (0,0)$ then $(c,d)$ can be $(0,15), (1,14), ...., (7,8)$ , giving us $8$ quadruples.

(ii) If $(a,b) = (0,1)$ then $(c,d)$ can be $(1,13), (2,12), ... ,(7,7)$ , giving us $7$ more quadruples.

And continuing with this brute force approach .....

(iii) $(a,b) = (0,2) \Longrightarrow (c,d) = (2,11), (3,10), ..., (6,7) \Longrightarrow 5$ more quadruples.

(iv) $(a,b) = (0,3) \Longrightarrow (c,d) = (3,9), (4,8), (5,7), (6,6) \Longrightarrow 4$ more quadruples.

(v) $(a,b) = (0,4) \Longrightarrow (c,d) = (4,7), (5,6) \Longrightarrow 2$ more quadruples.

(vi) $(a,b,c,d) = (0,5,5,5)$ for $1$ more quadruple.

(vii) $(a,b) = (1,1) \Longrightarrow (c,d) = (1,12), (2,11), ... ,(6,7) \Longrightarrow 6$ more quadruples.

(viii) $(a,b) = (1,2) \Longrightarrow (c,d) = (2,10), .... ,(6,6) \Longrightarrow 5$ more quadruples.

(ix) $(a,b) = (1,3) \Longrightarrow (c,d) = (3,8), (4,7) ,(5,6) \Longrightarrow 3$ more quadruples.

(x) $(a,b) = (1,4) \Longrightarrow (c,d) = (4,6), (5,5) \Longrightarrow 2$ more quadruples.

(xi) $(a,b) = (2,2) \Longrightarrow (c,d) = (2,9), ..., (5,6) \Longrightarrow 4$ more quadruples.

(xii) $(a,b) = (2,3) \Longrightarrow (c,d) = (3,7), (4,6), (5,5) \Longrightarrow 3$ more quadruples.

(xiii) $(a,b,c,d) = (2,4,4,5)$ for $1$ more quadruple.

Finally, we have $(3,3,3,6), (3,3,4,5)$ and $(3,4,4,4)$ , giving us $3$ more quadruples.

Adding up all the case results, we end up with a total of $\boxed{54}$ ordered quadruples.