The proof of the asymptotic formula
$$
\Phi(n) = \frac{3}{\pi^2} n^2 + O(n\log n)
$$
can be modified to give a simple upper bound that has the right asymptotic character (but is not very accurate). We'll follow the proof in Chandrasekharan's Introduction to Analytic Number Theory.
By interpreting $\Phi(n)$ as the number of lattice points with relatively prime coordinates in or on the triangle $0 < y \leq x \leq n$, Chandrasekharan obtains the formula
$$
\Phi(n) = \frac{\Psi(n)+1}{2},
$$
where
$$
\Psi(n) = \sum_{1 \leq d \leq n} \mu(d) \left\lfloor \frac{n}{d} \right\rfloor^2.
$$
If we write
$$
\left\lfloor \frac{n}{d} \right\rfloor = \frac{n}{d} - \left\{\frac{n}{d}\right\},
$$
then
$$
\begin{align}
\Psi(n) &= \sum_{1 \leq d \leq n} \mu(d)\left(\frac{n}{d} - \left\{\frac{n}{d}\right\}\right)^2 \\
&= n^2 \sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} - 2n \sum_{1 \leq d \leq n} \frac{\mu(d)}{d}\left\{\frac{n}{d}\right\} + \sum_{1 \leq d \leq n} \mu(d) \left\{\frac{n}{d}\right\}^2.
\end{align}
$$
Now
$$
\begin{align}
-2n \sum_{1 \leq d \leq n} \frac{\mu(d)}{d}\left\{\frac{n}{d}\right\} &< 2n \sum_{1 \leq d \leq n} \frac{1}{d} \\
&< 2n \left(1+\int_1^n \frac{du}{u}\right) \\
&= 2n(1+\log n)
\end{align}
$$
and
$$
\sum_{1 \leq d \leq n} \mu(d) \left\{\frac{n}{d}\right\}^2 < \sum_{1 \leq d \leq n} 1 = n,
$$
giving us
$$
\Psi(n) < n^2 \sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} + 2n\log n + 3n.
$$
For the sum we get
$$
\begin{align}
\sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} &= \sum_{1 \leq d \leq
\infty} \frac{\mu(d)}{d^2} - \sum_{n+1 \leq d \leq \infty} \frac{\mu(d)}{d^2} \\
&= \frac{6}{\pi^2} - \sum_{n+1 \leq d \leq \infty} \frac{\mu(d)}{d^2} \\
&< \frac{6}{\pi^2} + \sum_{n+1 \leq d \leq \infty} \frac{1}{d^2} \\
&< \frac{6}{\pi^2} + \int_n^\infty \frac{du}{u^2} \\
&= \frac{6}{\pi^2} + \frac{1}{n},
\end{align}
$$
so that, in total,
$$
\Psi(n) < \frac{6}{\pi^2} n^2 + 2n\log n + 4n
$$
and hence
$$
\Phi(n) < \frac{3}{\pi^2} n^2 + n\log n + 2n + \frac{1}{2}.
$$
$$\sum_{\substack{1\leq k\leq n\\(k,n)=1}} (n-k) = \sum_{\substack{1\leq k\leq n\\(k,n)=1}} k$$
Thus:
$$\begin{align}n\phi(n) &= n \sum_{\substack{1\leq k\leq n\\(k,n)=1}} 1\\ &= \sum_{\substack{1\leq k\leq n\\(k,n)=1}} n \\&= \sum_{\substack{1\leq k\leq n\\(k,n)=1}}(n-k) + \sum_{\substack{1\leq k\leq n\\(k,n)=1}} k \\&= 2\sum_{\substack{1\leq k\leq n\\(k,n)=1}} k\end{align}$$
Note, the first equality is only true for $n>1$, since when $n=1$ you have that $(1,1)=1$ so the left side is $1-1$ and the right side is $1.$
Best Answer
Compare the sum with the integral $$\int_Q^\infty \frac1{x^{2-\epsilon}}dx$$