Sum of magnitudes of coefficients of polynomial $(x-1)(x-2)(x-3)\cdots(x-(n-1))$

Hint: Let

$$f(x) = (x-1)(x-2)(x-3)...(x-(n-1)) \tag{1}\label{eq1A}$$

Note $f(-1) = (-2)(-3)\cdots(-n) = (-1)^{n-1}(n!)$. Consider what this value is in the expansion of $f(x)$ as a polynomial in $x$, plus how it relates to the sum of the magnitudes of the coefficients.

The only reason why you're adding up the magnitudes of the coefficients, and not the coefficients themselves, is that some of them are negative. If instead of considering $(x-1)(x-2)\dots(x-n)$ we consider $(x+1)(x+2)\dots(x+n)$, we'll be left with a polynomial whose coefficients have the same magnitudes, but are always positive. So let's do that.

Now, for any polynomial $P(x)$, $P(1)$ gives the sum of the coefficients: if $P(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$, then $P(1)=a_0+a_1(1)+a_2(1^2)+\dots+a_n(1^n)=a_0+a_1+a_2+\dots+a_n$.

So let $P_n(x)=(x+1)(x+2)\dots(x+n)$. Then

\begin{align*} P_n(1)&=(1+1)(1+2)\dots(1+n)\\ &=2(3)\dots(n+1)\\ &=(n+1)! \end{align*}

That is, the sum of the coefficients of $P_n$ is $(n+1)!$, as you've observed.

Since all roots of the polynomial are positive, Descartes' Rule of signs guarantees that the signs of the coefficients alternate, for example with $n=4$:

$(x-1)(x-2)(x-3)=x^3\color{purple}{-}6x^2\color{purple}{+}11x\color{purple}{-}6$

Because of this sign pattern the absolute values of the coefficients are recovered by putting $x=-1$ so that the powers of $x$ correlate with the alternating signs of the coefficients:

$(-1-1)(-1-2)(-1-3)=(-1)^3-6(-1)^2+11(-1)11x-6=-(1+6+11+6)$

and upon taking absolute values

$(1+1)(1+2)(1+3)=1+6+11+6$

where for the general case the product on the left is $n!$ and the sum on the right contains the desired coefficient magnitudes.