The $f$ you defined does not have this property when $n\ge 4$.
We can show that, if $(x_1,\dots,x_n)$ are arbitrary, when
$$t_\sigma=f(x_{\sigma(1)},\dots,x_{\sigma(n)})$$
then the set of images $T=\{t_\sigma:\sigma\text{ is a permutation}\}$ has maximum cardinality exactly $(n-1)!$.
Proof:
Write $f(x_1,\dots,x_n) = g(x_1,\dots,x_n)^n$. Note that a cyclic permutation of $(x_1,\dots,x_n)$ multiplies $g$ by $\alpha_n^k$ and therefore $f$ is multiplied by $\alpha_n^{kn} = 1$. So since $t_{\pi\circ\sigma}=t_\sigma$ when $\pi$ is one of the $n$ cyclic permutations, the maximum cardinality of $T$ is at most $n!/n=(n-1)!$.
Now, since $g$ is a linear function with distinct coefficients before each $x_i$, if you are working in a characteristic zero field, the set of all $g(x_{\sigma(1)},\dots,x_{\sigma(n)})$ has maximum cardinality $n!$. But since the equation $t_\sigma=X^n$ has at most $n$ solutions in $X$, the maximum cardinality of $T$ is at least $(n-1)!$, finishing the proof.
Now, why doesn't it work? In fact I don't know if the resolvent can be defined by a systematic formula that extends for all $n$: we just find ad-hoc polynomials that happen to let us reduce the equation to a degree as low as possible. As a result, proving that such a polynomial exists is much easier than proving that no polynomial exists, and I don't know if there are proofs of Abel-Ruffini that don't use group theory.
So, how could we define a resolvent for $n=4$? Since $n$ is even there are $\tbinom{n}{n/2}/2=3$ ways to partition the roots into a set of two equal-sized subsets, so that:
$$f(x)=x_1x_2+x_3x_4$$
has at most 3 images under permutation of $x$ (and it is easy to check that this bound is reached). There are 8 symmetries that can be derived from the two generator equations:
$$f(a,b,c,d)=f(b,a,c,d)\\
f(a,b,c,d)=f(c,d,a,b)$$
Note that in general $f(a,b,c,d)\ne f(b,c,d,a)$, so unlike when $n=3$ cylic permutations are not always symmetries. Also note that this construction is only really useful for $n=4$, since for $n=6$ we get $\tbinom{n}{n/2}/2=10>6$.
You can check Lagrange's original 1770 paper (in French) to see that it is indeed the way he defined cubic resolvents and quartic resolvents.
He also gives a much more complicated polynomial for quintics, which he derives by generalizing to any root of unity $\alpha$ ($\alpha^5=1$), expanding the $f$ you just defined as a linear combination of $\alpha^0,\dots,\alpha^{n-1}$ and taking only the coefficient of $\alpha^0$. It turns out the resulting expression is highly symmetric, and the 20 symmetries can be derived from the three generators:
$$\begin{align}
f(a,b,c,d,e)&=f(b,c,d,e,a)&&\text{cyclic permutation}\\
f(a,b,c,d,e)&=f(e,d,c,b,a)&&\text{reversal}\\
f(a,b,c,d,e)&=f(a,c,e,b,d)&&\text{doubling modulo 5}
\end{align}$$
so that there are $5!/20=6$ distinct values. Cyclic permutation is for the same reason as when $n=3$; reversal and doubling modulo 5 come from the fact when replacing $\alpha$ by another root of unity $\alpha'=\alpha^k$ (where $k$ is relative prime with $n$), the coefficient of $\alpha'^0$ is the same as that of $\alpha^0$ so that $f$ is unchanged. In fact reversal is even implied by doubling modulo 5.
This construction of course generalize to all prime $n$, and in fact we have the closed form
$$f(x_1,\dots,x_n)=\sum_{i_1+\dots+i_n=0\pmod n} x_{i_1}\dots x_{i_n}$$
which has $n(n-1)$ symmetries corresponding to the invertible affine maps $z\mapsto az+b$ of $\mathbb Z/n\mathbb Z$ (such maps preserve the condition $i_1+\dots+i_n=0 \pmod n$), so that the maximum cardinality of the images of $f$ is at most $(n-2)!$ (so for $n=3$, this degenerates to a completely symmetric function that is of no use to actually solving the equation because it doesn't give us any relation between the roots).
Best Answer
Notice that $V(\alpha)$ is quadratic in $\alpha$ - you have an $\alpha^{2}$ term, a $(1-\alpha)^{2}=\alpha^{2}-2\alpha+1$ term, and a $\alpha(1-\alpha)=\alpha-\alpha^{2}$ term. So you can complete the square, then set the squared term to zero. However, this will only give you a minimum if the squared term's coefficient is positive.
You can also solve it via calculus, like DMcMor did. In general calculus is the way to go, but for quadratics completing the square is a nice alternative.
Answer in spoilers:
EDIT:
Since OP has stated $\alpha$ is restricted to the interval $[0,1]$ (correct me if the interval is not closed), it might be the case that the minimum obtained by either this method or calculus will lie outside that interval. In this case, one would pick the value that is as close as possible to that minimum - say you got $1.5$, you would choose $\alpha=1$. If you got $-0.5$, you'd pick $\alpha=0$. You can do this because a quadratic is monotonic either side of its turning point. In fact, if you don't find any turning points in the allowed interval, the function must be monotonic if it is continuous, so the minimum in that interval must be one of the two endpoints (and it's usually simple to determine which).