I'm trying to show that $\bar{p}$ is a minimum
Writing the function as $$f(p) =\sum_{i=1}^n\langle p-p_i,p-p_i\rangle,$$we have $$Df(p)(v) =\sum_{i=1}^n 2\langle p-p_i,v\rangle.$$So, write this as $$Df(p)(v) =\left\langle 2np-\sum_{i=1}^n2p_i,v \right\rangle$$to conclude that $$\nabla f(p) = 2np-\sum_{i=1}^n2p_i. $$ So $\nabla f (p) $ is the zero vector if and only if $$p= \frac{1}{n}\sum_{i=1}^n p_i. $$To see that this gives the minimum, compute the Hessian: $$D^2f(\overline{p})(v,w) = C\langle v,w\rangle $$for some $C>0$. Since $D^2f (\overline{p})$ is positive-definite, we are done.
For the part $$Df(p)(v) =\left\langle 2np-\sum_{i=1}^n2p_i,v \right\rangle$$How do we conclude that $$\nabla f(p) = 2np-\sum_{i=1}^n2p_i$$since we can't divide by a vector?
There is an integral Theorem in Analysis linking the directional derivative to the derivative. To be more precise:
Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ be continous differentiable in $x_0$. The directional derivative $Df(x_0)(v)$ exists for every $v \in \mathbb{R}^n \setminus \{0\}$ and one has $Df(x_0)(v) = \langle \nabla f(x_0),v \rangle$.
So for your problem you immediately get that $\nabla f(p) = 2np-\sum_{i=1}^n2p_i$