First of all, the integer solutions of $x_1+x_2+x_3+x_4=n$ where $0\leq x_1,$ $2\leq x_2,$ $x_3$ is a multiple of $4$, and $0\leq x_4\leq 3$ can be represented by: \begin{align} x_1:&\ (1+x+x^2+x^3+\dots)=\frac{1}{1-x}\\ x_2:&\ (x^2+x^3+x^4+\dots)=\frac{x^2}{1-x}\\ x_3:&\ (1+x^4+x^8+x^{12}+\dots)=\frac{1}{1-x^4}\\ x_4:&\ (1+x+x^2+x^3)\\ \end{align} which means the generating function for this equation is: $$ G(x)= \frac{x^2(1+x+x^2+x^3)}{(1-x)^2(1-x^4)} $$ sidenote: is it possible to write this summation form?
Next, I wish to find a general expression for the inequality $x_1+x_2+x_3+x_4\leq n$, but I'm unsure how to use the generating function above to do so, or if I need it at all.
To compute the generating function of the partial sums, multiply by $1/(1-x)$, yielding $$H(x) = \frac{x^2(1+x+x^2+x^3)}{(1-x)^3(1-x^4)}$$
To find an explicit formula from the generating function, use partial fraction decomposition: \begin{align} H(x) &= \frac{1}{(1-x)^2} - \frac{2}{(1-x)^3} + \frac{1}{(1-x)^4}\\ &= \sum_{n \ge 0} \binom{n+1}{1} x^n - 2 \sum_{n \ge 0} \binom{n+2}{2} x^n + \sum_{n \ge 0} \binom{n+3}{3} x^n\\ &= \sum_{n \ge 0} \left(\binom{n+1}{1} - 2 \binom{n+2}{2} + \binom{n+3}{3}\right) x^n, \end{align} which immediately implies that the number of solutions is $$\binom{n+1}{1} - 2 \binom{n+2}{2} + \binom{n+3}{3} = \frac{n^3-n}{6}$$