Welcome to the amazing world of generating functions. Firstly, while the polynomials you set up are correct, as you will see it's more convenient to use infinite series instead of the truncated series as you have done, meaning to instead analyse the power series
\begin{align*}
G(x) &= (x + x^2 + \cdots + x^7)(x^9 + x^{10} + \cdots)(x + x^2 + \cdots)^3 \\
&= x^{13}\underbrace{(1 + x + \cdots + x^6)(1 + x + x^2 + \cdots)^4} \\
&= x^{13}H(x)
\end{align*}
Now we want to find $[x^{29}]G(x) = [x^{16}]H(x)$. Firstly, we transform these ugly series into a closed form. We have a beautiful tool called geometric series for this!
$$
H(x) = \left(\sum_{i = 0}^6 x^i\right)\left(\sum_{i = 0}^{\infty} x^i\right)^4 = \frac{1 - x^7}{(1 - x)^5}
$$
The next natural step is to simplify the fraction, since the degree of the numerator exceeds the denominator:
$$
H(x) = -\frac{35x^4 - 105x^3 + 126x^2 - 70x + 14}{(1 - x)^5}
$$
[Note 1]
Now, the remaining step is quite simple. We simply expand out $\frac{1}{(1 - x)^5}$ into a power series again. There are many ways to figure this out, but the standard way is to take repeated derivative of $\frac{1}{1 - x} = \sum_i x^i$. You can check out this answer for more details, but I will give the answer here - we have
$$
\frac{1}{(1 - x)^5} = \sum_{k = 0}^{\infty} \binom{k + 4}{4} x^k = 1 + 5x + 15x^2 + 35x^3 + \cdots
$$
Finally, we can substitute this into our $H(x)$ to get
$$
H(x) = (\color{red}{-35x^4} + 105x^3 - 126x^2 + 70x - 14)\sum_{k = 0}^{\infty} \binom{k + 4}{4} x^k
$$
You can write this into a single summation (without the multiplication) by multiplying the sums out and shifting indices around, but I won't bother, since our final goal is to find $[x^{16}]H(x)$, which we can do here:
$$
[x^{16}]H(x) = \color{red}{-35[x^{12}]}\sum_k \binom{k + 4}{4} x^k + \cdots = -35 \cdot \binom{16}{4} + \cdots
$$
I am feeling lazy so I won't write everything out, but that's the idea. Also, here's a way to check the answer using Sage, a computer algebra system. The answer should be 4130 :)
Oh and as a bonus, you can write down an explicit answer for any coefficient, meaning you can answer the question not just for $29$ but for any number! You can also derive recurrence relations between the coefficients by dealing with the polynomial itself (c.f. Fibonacci numbers here), but you can explore that yourself.
[Note 1]: Normally, here you would have to first do a partial fraction decomposition. For example, suppose that one of the constraints is that $a$ or any of the variable has to be even. In that case, your power series would have something of the form $1 + x^2 + x^4 + \cdots$, meaning $H(x)$ will then look like $\frac{\textrm{polynomial}}{(1 - 2x)(1 - x)^4}$. You perform partial fraction decomposition to get $\frac{\cdot}{1 - 2x} + \frac{\cdot}{1 - x} + \cdots$, then apply the same method.
