By translations $x\to x+c$ and symmetries $x\to-x$ you can reformulate your problem as
Find all $x_1,x_2,x_3,x_4,x_5$ integers that satisfy
$$
1+x_3+x_4 = x_1+x_2+x_5$$
with bounds
$$0\le x_1\le 5\qquad 0\le x_2\le 5\qquad 0\le x_5\le 6$$
$$0\le x_3\qquad 0\le x_4$$
Now, call $p(n)$ the number of triples $(x_1,x_2,x_5)$ such that $x_1+x_2+x_5=n$ and such that
$$0\le x_1\le 5\qquad 0\le x_2\le 5\qquad 0\le x_5\le 6$$
hold. In the same way, let $q(n)$ the number of couples $(x_3,x_4)$ such that $x_3+x_4=n$ and such that
$$0\le x_3\qquad 0\le x_4$$
hold. Your solution is
$$
p(1)q(0) + p(2)q(1) + p(3)q(2) + \dots + p(16)q(15).
$$
We know that $q(n)=n+1$, but $p(n)$ is much harder to compute, and I think must be done by a computer.
Let's do it with permutations/inclusion-exclusion.
First of all, a Toy Problem
Find all $x_1,x_2,x_3,x_4,x_5$ positive integers that satisfy
$$c+x_3+x_4 = x_1+x_2+x_5$$
where $c$ is an integer (may be negative) and with bound
$$c + x_3+x_4\le n$$
where $n$ is a positive integer.
Like before, let $p(m)$ the number of triples $(x_1,x_2,x_5)$ such that $x_1+x_2+x_5=m$ and let $q(m)$ the number of couples $(x_3,x_4)$ such that $x_3+x_4=m$. We know that
$$
p(m) = \frac{(m+2)(m+1)}2 = \binom{m+2}2 \qquad q(m) = m+1 = \binom{m+1}1
$$
The solution is
$$TP(c,n) = q(-c)p(0) + q(1-c)p(1) + \dots +q(n-c)p(n) $$
$$ = \binom {1-c}1\binom 22 + \binom {2-c}1\binom 32 + \dots + \binom {n-c+1}1\binom {n+2}2 $$
If we return to our first problem, we can operate an inclusion-exclusion method.
- First of all, we notice that $x_1+x_2+x_5\le 16$, so we can consider the Toy Problem with $c=1$ and $n=16$, so we get $TP(1,16)$.
- We have to respect the bounds, though, so we have to subtract the wrong solutions. How many solution are there with $x_5>6$? If we substitute $x_5 = 7 +z_5$ we get again the Toy Problem with
$$-6+x_3 + x_4 = x_1+x_2+z_5$$
so $c=-6$ and $n=16 -7=9$, so $TP(-6,9)$. If we do the same for $x_1,x_2$ we get $TP(-5,10)$ twice.
- Now we subtracted too much. For example we subtracted twice the (wrong) solutions with $x_1>5,x_2>5$ and we have to add them again. With the same substitutions as before, we get $TP(-11,4)$. Doing the same for the other couples, we get $TP(-12,3)$ twice.
The final answer is thus
$$TP(1,16) - TP(-6,9) - 2\cdot TP(-5,10) + TP(-11,4) + 2\cdot TP(-12,3) $$