jmc

The degree of the polynomial is $1+2+3+\cdots +12=\frac{12\cdot 13}{2}=78.$

When we expand $(x-1)(x^2-2)(x^3-3)\cdots (x^{11}-11)(x^{12}-12),$ we choose a term from each factor. For example, from the first factor $x-1,$ we can choose either $x$ or $-1.$ From the second factor $x^2-2,$ we can choose either $x^2$ or $-2,$ and so on. So to find the coefficient of $x^{70},$ we want to cover all possible choices where the powers of $x$ multiply to $x^{70}.$

Since the degree of the polynomial is $x^{78},$ the product of the "missing" powers of $x$ must be $x^8.$ We divide into cases.

Case 1: One factor has a missing power of $x.$

If one factor has a missing power of $x,$ it must be $x^8-8,$ where we choose $-8$ instead of $x^8.$ Thus, this case contributes $-8x^{70}.$

Case 2: Two factors have a missing power of $x.$

If there are two missing powers of $x,$ then they must be $x^a$ and $x^b,$ where $a+b=8.$ The possible pairs $(a,b)$ are $(1,7),$ $(2,6),$ and $(3,5)$ (note that order does not matter), so this case contributes $[(-1)(-7)+(-2)(-6)+(-3)(-5)]x^{70}=34x^{70}.$

Case 3: Three factors have a missing power of $x.$

If there are three missing powers of $x,$ then they must be $x^a,$ $x^b,$ and $x^c,$ where $a+b+c=8.$ The only possible triples $(a,b,c)$ are $(1,2,5)$ and $(1,3,4),$ so this case contributes $[(-1)(-2)(-5)+(-1)(-3)(-4)]x^{70}=-22x^{70}.$

Case 4: Four factors or more have a missing power of $x.$

If there are four or more missing powers of $x,$ then they must be $x^a,$ $x^b,$ $x^c,$ and $x^d$ where $a+b+c+d=8.$ Since $a,$ $b,$ $c,$ $d$ are distinct, we must have $a+b+c+d\ge 10.$ Therefore, there are no ways to get a power of $x^{70}$ in this case.

Thus, the coefficient of $x^{70}$ is $(-8)+34+(-22)=\boxed{4}.$