You've shown that $f(a)\equiv0\pmod{2}$ for $a=1$ and $a=2$. To show that there exists some $\alpha\in\Bbb{Z}_p$ such that $f(\alpha)=0$, you can apply Hensel's lemma as you quote it in your question; it suffices to show that $f'(a)\not\equiv0\pmod{2}$ for some $a$.
Lemma: Let $p \neq 2$ be a prime, $a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p|a$, and $Q: ax^2 + by^2 + cz^2 = 0$ a quadratic form. Then there is a nontrivial solution to $Q$ over $\mathbb{Q}_p$ if and only if $-b/c$ is a square mod $p$.
Proof: Suppose a solution exists. Then by scaling, we may assume that $y$ and $z$ lie in $\mathbb{Z}_p$, and indeed that they lie in $\mathbb{Z}_p^\times$. (Suppose without loss of generality that $y \in p\mathbb{Z}_p;$ then as $p|a$, $p|y$ and $p\nmid c$, we must have $p|z,$ and hence $p|x$ by considering the parity of the exponent of $p$ in the sum. So we can divide our trio $(x,y,z)$ by $p$.)
Now consider the equation mod $p$. This becomes equivalent to $(y/z)^2 \equiv -b/c$ mod $p,$ where we can divide by $c$ and $z^2$ as they have invertible image in $\mathbb{F}_p$. So $y/z$ gives the corresponding element of $\mathbb{F}_p$ that squares to $-b/c$.
Conversely, suppose that we have $Y^2 \equiv -b/c$ mod $p$. Then in particular, $p\nmid Y$ and the triple ($x,y,z) = (0,Y,1)$ gives a solution mod $p$. Using Hensel's lemma (and using that $p\neq 2$) we see that this lifts to a solution in $\mathbb{Q}_p$, as required.
Reference: At which p-adic fields does the equation have no solution?
Best Answer
For $n\equiv 1,2,3,5,6\pmod8$ there's an integer solution to $x^2+y^2+z^2\equiv n\pmod 8$ with $x$ odd. Then $n-y^2-z^2$ is an integer congruent to $1$ modulo $8$, and so is a $2$-adic square, $x'^2$. Then $x'^2+y^2+z^2=n$ over $\Bbb Z_2$.
If $4\mid n$ consider $n/4$ etc.