Since for all $x \in \mathbb{R}$, $(2^{1+x}+2^{1-x}),\ f(x),\ (3^x+3^{-x})$ are in A.P., the middle term equals the average of the other two terms.

Hence, $f(x)=\frac{(2^{1+x}+2^{1-x})+(3^x+3^{-x})}{2}.$

Simplify the first part: $2^{1+x}+2^{1-x}=2\cdot 2^x+2\cdot 2^{-x}=2\left(2^x+2^{-x}\right).$

Therefore, $f(x)=\frac{2(2^x+2^{-x})+(3^x+3^{-x})}{2}$ $=\left(2^x+2^{-x}\right)+\frac{1}{2}\left(3^x+3^{-x}\right).$

Now use the standard inequality: $a+\frac{1}{a} \ge 2 \quad \text{for } a>0,$ with equality when $a=1$.

Since $2^x>0$ and $3^x>0$, we get $2^x+2^{-x} \ge 2,$ $3^x+3^{-x} \ge 2.$

Substitute these into $f(x)$: $f(x) \ge 2+\frac{1}{2}(2)=2+1=3.$

Check equality: Equality holds when $2^x=1 \quad \text{and} \quad 3^x=1,$ which happens at $x=0.$

Then $f(0)=3.$

Therefore, the minimum value of $f(x)$ is $\boxed{3}.$

Option check: