I finally came up with a solution, which seems to be a valid reformulation.
Expand $f_i(x) \ (i=0,1,\cdots,m)$ into the form (form clarity, I drop the subscript in $f_i(x), P_i$ and $q_i$.
$$f(x)=\frac{1}{2}\sum_i^{n}P_{ii}x_i^2+\sum_{i\neq j}P_{ij}x_i x_j+ \sum_i^n q_i x_i + r$$
Notice that $x \geq 0$, apply the change of variable $y_i=x_i^2$, we have
$$f(y)=\frac{1}{2}\sum_i^{n}P_{ii}y_i+\sum_{i\neq j}P_{ij}\sqrt{y_i y_j}+ \sum_i^n q_i\sqrt{y_i} + r$$
The first term $\sum_i^{n}P_{ii}y_i$ is affine in $y_i$. What is more, since the off-diagonal entries of $P$ are non-positive and this makes the second term convex. Similarly, the last term $\sum_i^n q_i\sqrt{y_i} + r$ is also convex. Then $f(y)$ is convex.
The "hint" directly from the problem is provided by $x\geq 0$, what indicates the possibility of taking the square root.
Notice that $x^2-1$ is a convex function. If we consider $x^2-1 \ge 0$, we have $x \ge 1$ or $x \le -1$. That is the feasible set is not a convex set.
However, notice that $x^2-1 \le 0$ is equivalent to $-1 \le x \le 1$ is convex.
In general, we know that $\{ x \mid f_i(x) \le 0\}$ is a convex set and their intersection, that is the feasible set that you have written down is a convex set. It is a desirable property to minimize a convex objective function over a convex set, in particular, we know that a local minimum is a global minimum.
Also, notice that the first example is a special case of the general form.
$$\begin{align} \text{max } x_1 & +5 x_2 \\
\text{s.t. } x_1 & \le 150 \\
x_2 &\le 350 \\
x_1+x_2 &\le 400 \\
x_1 , x_2 &\ge 0
\end{align}$$
Is actually equivalent to
$$\begin{align}- \min -x_1 & -5 x_2 \\
\text{s.t. } x_1 & \le 150 \\
x_2 &\le 350 \\
x_1+x_2 &\le 400 \\
-x_1 &\le 0 \\
-x_2 &\le 0
\end{align}$$
Here $f_0$ is just $-x_1-5x_2$ and the $f_i$ are just the LHS of the constraint.
We can flip the inequality by multiplying a negative sign and in fact the general form even include the first form. The general form doesn't tell us whether $x_i$ is bounded above or below since linear functions are convex and we can flip the inequality signs. The crucial propery here is convexity.
For the follow up question:
Even if $f_i$ is not convex, it is still possible for the feasible region to be bounded. In fact, in special cases, it is even possible for it to be convex. As an example
Consider $$x^3-1 \le 0$$
Even thought the function $x^3-1$ is not convex, the feasible region is just $x \le 1$.
Imposing $f_i$ to be convex explicitly gives us a convenient way to construct a convex set and also use its properties in deriving algorithms or investigate property of this form of problems.
Best Answer
The problem is nonconvex both from the indefinite quadratics, and from the combinatoric $x$. The way I would approach it is by first writing the absolute values as $-t \leq x^TQ_ix \leq t$ with objective $t$, and then exactly linearize the quadratic terms.
Since I am lazy, I will write the linearization for binary $x$ (scale and shift accordingly for your case). Terms $x_ix_j$ are replaced with $w_{ij}$ which satisfies $w_{ij}\leq x_i$, $w_{ij}\leq x_j$ and $w_{ij} \geq x_i + x_j-1$. At this point you have a standard mixed-integer linear program in $x$, $w$ and $t$.