What is the coefficient of the $x^3$ term in the expansion of $(x^2+x-5)^7$ (See details)?

Well, in full generality, the approach you took is the right one (though you made a minor mistake in the end), but if you want to do without all the machinery, it's probably easier to do this calculation armed only with a bit of common sense.

You want three factors of $x$ – where are you going to get them from? Either from three individual $x$s, or from one $x^2$ and one $x$. In either case, the remaining factors have to be $-5$. In the first case, you can choose $3$ out of $7$ factors to be $x$, so this yields $\binom73(-5)^4$. In the second case, you can select one out of $7$ factors to be $x^2$ and then one of the $6$ remaining factors to be $x$, so this yields $7\cdot6(-5)^5$. The sum is $35(-5)^4+42(-5)^5=(35-210)(-5)^4=-175\cdot625=-109375$, which is also what your approach would have yielded if you hadn't accidentally increased the exponent of $-5$ by one in the second term.

Why not use Calculus? Calculate the third derivative, evaluate at $0$, and divide by $3!=6$. Since the third derivative of $g^7$ is $210g^4{g'}^3+126g^5g'g''+7g^6g'''$, and $g(0)=-5$, $g'(0)=1$, $g''(0)=2$, and $g'''=0$, the evaluation gives surenough $-109375$.