Note that each given equation is of the form $$f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7$$ for some $k\in\{1,2,3\}.$
When we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ $$f(k)=ak^2+bk+c,$$ where $a,b,$ and $c$ are linear combinations of $x_1,x_2,x_3,x_4,x_5,x_6,$ and $x_7.$
We are given that
$$\begin{alignat*}{10} f(1)&=\phantom{42}a+b+c&&=1, \\ f(2)&=4a+2b+c&&=12, \\ f(3)&=9a+3b+c&&=123, \end{alignat*}$$
and we wish to find $f(4).$
We eliminate $c$ by subtracting the first equation from the second, then subtracting the second equation from the third:
$$\begin{align*} 3a+b&=11, \\ 5a+b&=111. \end{align*}$$
By either substitution or elimination, we get $a=50$ and $b=-139.$ Substituting these back produces $c=90.$
Finally, the answer is $$f(4)=16a+4b+c=\boxed{334}.$$
~Azjps ~MRENTHUSIASM