Let $v$ denote the $p$-adic valuation in $\Bbb{Z}_p$. Let $a_1,a_2,a_3$ be three elements of $\Bbb{Z}_p$. Then I want to show that
$$\min \{ v(a_1), v(a_2), v(a_1+a_2+a_3)\}=\min \{v(a_1),v(a_2),v(a_3)\}$$
I proved this when we have two elements but I cannot prove this with three elements. Does it hold if we take $n$ number of elements? Thank you in advance
Suggestion: Use your $2$-element version to show:
$$\min\{v(a),v(b),v(c)\}=\min(v(a),v(b),v(a+c)\}$$
Then apply this result twice.
You will be using $\min(x,y,z)=\min(\min(x,z),y)$.
At heart, note that the "min" binary operator is associative and commutative, and $\min(x_1,x_2,\dots,x_n)$ is just the application of this binary operator $n-1$ times, in any order on this list of elements.
And yes, it will generalize to $n$ elements.
Verbose argument:
If $x\star y = \min(x,y)$, then you already have:
$$v(a)\star v(b)=v(a)\star v(a+b)\tag{1}$$
Now:
$$\begin{align} v(a)\star v(b)\star v(c) &= v(a)\star\left(v(b)\star v(b+c)\right)\\ &=\left(v(a)\star v(b+c)\right)\star v(b)\\ &=(v(a)\star v(a+b+c))\star v(b)\\ &=v(a)\star v(b)\star v(a+b+c) \end{align} $$
Essentially, associativity and commutativity of $\star$ lets are re-arrange, apply $(1)$ rearrange again, apply (1) again, and re-arrange a last time.