Given $$\models\phi(p_0,\dots p_n)$$
Let $\psi_1, \dots \psi_n$ be sentences.
Prove that
$$\models \phi(\psi_1, \dots \psi_n)$$
My proof:
Let $\pi(x)$ denote the logical value of sentence $x$. Then, $\pi(\phi(\psi_1, \dots, \psi_n)) = \phi(\pi(\psi_1), \dots, \pi(\psi_n)) $
And since $\phi$ is a tautology, it is true for all possible combinations of input values, therefore
$$\models \phi(\psi_1, \dots, \psi_n)$$
Is this a valid way to solve this problem?
More or less, yes. I recommend distinguishing the $\phi$ in $\phi(\psi_1,\dots,\psi_n)$ and the $\phi$ in $\phi(\pi(\psi_1),\dots,\pi(\psi_n))$. One is a well-formed formula into which you are substituting several other formulas. The other is a Boolean function which you are applying to $n$ Boolean values. You could use a notation like $\phi_\pi(\pi(\psi_1),\dots,\pi(\psi_n))$, but I like using an outfix notation $[\![\phi]\!]$ for $\pi$, and I would write $[\![\phi(\psi_1,\dots,\psi_n)]\!]=[\![\phi]\!]([\![\psi_1]\!],\dots,[\![\psi_n]\!])$.
Next, this is minor, but you should say why the previous equation holds. In this case, it's by definition of interpretation. That is, not just any function $\pi$ is allowed as an appropriate notion of "logical value", to use your words, but only ones that satisfy the given equation.
Finally, if you wanted to be extra formal, you might do an induction on the number of free variables in $\phi$, i.e. on $n$.