Positive harmonic function on $\mathbb{R}^n$ is a constant? Is it true that a positive harmonic function on $\mathbb{R}^n$ must be a constant? How might we show this? The mean value property seems not to be the way...for that we would need boundedness.

Taking $n=2$ just to simplify:
Note that, given $y\in\mathbb{R}^2$, let $r=||y||$. Then, $\forall n\in\mathbb{N}$:
$$B(y,n)\subset B(0,2n)$$
Thus
$$\int_{B(y,n)}u(z)dz\leq\int_{B(0,n+r)}u(z)dz$$
$$u(y) = \frac{1}{\pi n^2}\int_{B(y,n)}u(z)dz\leq\frac{(n+r)^2}{n^2}\frac{1}{\pi(n+r)^2}\int_{B(0,n+r)}u(z)dz=\left(\frac{n+r}{n}\right)^2u(0)$$
That is 
$$u(y)\leq\left(\frac{n+r}{n}\right)^2u(0),\qquad\forall n \in\mathbb{N}$$
Takin $lim_n\rightarrow \infty$, we have that $u(y)\leq u(0)$. But since $y\in\mathbb{R}^2$ was arbitrary, we conclude that $u(z)\leq u(0),\forall z\in\mathbb{R}^2$.
But then, given $R>0$, we have that $\max_{z\in\overline{B(0,R)}}u(z) \leq u(0)$, thus, $\max_{z\in\overline{B(0,R)}}u(z) = u(0)$. Since $0$ is an interior point, by the strong maximum principle, $u$ is constant equal to $u(0)$ in the ball. Since $R$ was arbitrary, it follows that $u(z) = u(0),\forall z\in\mathbb{R}^2$.
