Is there a nice explanation for this curious fact about cyclic subgroups?

(Essentially the same answer as Neil Strickland's:) Since a cyclic group of order $n$ has $\varphi(n)$ generators, your sum equals $$ \DeclareMathOperator{\ord}{ord} \nu(G) = \sum_{g\in G} \frac{ \ord(g) }{ \varphi(\ord(g)) } . $$ For elements $g$ of $p$-power order, we get $\ord(g)/ \varphi( \ord(g) ) = p/(p-1)$ and thus if $G$ is a $p$-group, $\nu(G) = 1 + (|G|-1)p/(p-1)$ depends only on the order of $G$ and not on its structure, and so $\nu(G) =\nu(C_{|G|})$. So your equality holds for nilpotent groups, as was remarked in Nilpotency of a group by looking at orders of elements .
See also the papers http://arxiv.org/abs/1207.1020 and http://arxiv.org/abs/1503.00355.

This may well be the proof that you have already.

Reduce to the case of $p$-groups as in the question. Put \begin{align*} n(x) &= |\langle x\rangle| \\ m(x) &= |\{x' : \langle x'\rangle = \langle x\rangle\}| \end{align*} Then $$ \nu(G) = \sum_{H \text{cyclic}} |H| = \sum_{x\in G} n(x)/m(x). $$ Now, if $x=1$ then $m(x)=1$. On the other hand, if $x\neq 1$ then $n(x)=p^k$ for some $k>0$ and $m(x)=p^k-p^{k-1}=(1-1/p)n(x)$. This gives \begin{align*} \nu(G) &= 1 + \sum_{x\neq 1} \frac{1}{1-1/p} = 1 + \frac{|G|-1}{1-1/p} \end{align*} If $|G|=p^r$, this becomes $$ \nu(G) = 1 + \frac{p^r-1}{1-1/p} = \frac{p^r-1/p}{1-1/p} = \frac{p^{r+1}-1}{p-1} = \sum_{i=0}^r p^i = \sigma_1(p^r). $$

Another proof of Strickland's result.

Let $G$ be a group of order $p^n$ and $\nu(G)$ be the sum of orders of its cyclic subgroups. To prove that $\nu(G)=\sigma_1(G)$, we proceed by induction on $|G|$. If $G$ is cyclic, then, obviously, $\nu(G)=\sigma_1(G)$. Therefore assume that $G$ is noncyclic. If $H<G$, then, by induction, $\nu(H)=\sigma_1(H)$ depends only on $|H|$. Therefore assume that $G$ is noncyclic. Then it has a normal subgroup $T$ such that $G/T$ is elementary abelian of order $p^2$. In that case, as all maximal subgroups of $G$ have the same order, we get $$ \nu(G)=\sum_{T<H<G}\nu(H)-p\nu(T)=(p+1)\nu(H)-p\nu(T) $$ hence, by induction, $$ \nu(G)=(p+1)\sigma_1(p^{n-1})-p\sigma_1(p^{n-2}) $$ $$ =(p+1)(1+p+\dots+p^{n-1})-p(1+p+\dots+p^{n-2})=1+p+\dots+p^n=\sigma_1(p^n), $$ completing the proof.

Above we applied a partial case of Hall's enumeration principle. Applying that principle, it is easy to compute, for not very complicated $p$-groups, the sum of orders of their subgroups. About Hall's enumeration principle, see Kap. III in Huppert's `Endliche Gruppen'.