TLDR
Comparing the magnetic field strength of a permanent magnet with that of a planet is naive. Such a comparison ignores the fact that magnetic fields dissipate with distance much as gravity does. If you compare their magnetic properties independent of distance, their magnetic moment in particular, you'll see that Earth's magnetic moment is 75 trillion times more powerful than industrial grade rare-earth magnets which simply outclasses permanent magnets entirely. Artificially generating a similar field for Mars requires similarly powerful engineering, or equally brilliant idea.
Permanent Magnets
Permanent magnets have a remanence (magnetic strength, $B_r$) of $0.2$ $T$ to $1.4$ $T$ depending on the composition. Ferrite magnets have a $B_r$ of $0.2$ $T$ - $0.4$ $T$ whereas sintered neodymium rare earth magnets have a $B_r$ of $1.0$ $T$ - $1.4$ $T$ [Wikipedia]
Earth
Earth's surface [magnetic field strength] ($B_s$) ranges from 25 to 65 microteslas ($25 \times 10^{-6}$ $T$ - $65 \times 10^{-6}$ $T$).
Relating the Two
Disclaimer: Please bear with me here. EM was not my strong suit.
Your source suggests the usage of permanent magnets because, on face value, permanent magnets appear stronger than planetary magnetic fields. Using the above values, one would think that permanent magnets are 26,000 to 17,000x more powerful than Earth's magnetic field. The naivete of this notion is that it ignores that magnetic field strength decreases with distance. In fact it does so similarly to gravity: $B \propto {{1}\over{r^2}} $, $G \propto {{1}\over{r^2}} $.
A better metric to compare the two 'magnets' would be to compare their magnetic moments. Magnetic moment is normally defined as a magnet's responsiveness to an external magnetic field; but, it's also used to define the magnet's magnetic field when it is known.
Permanent Magnet's Magnetic Moment
For permanent magnets, their magnetic moment is simply defined as:
$\vec{m} = {{1}\over{\mu_0}} \vec{B_r} V$
where $\mu_0$ is the vacuum permeability constant, $\vec{B_r}$ is the remanence (magnetic field strength) of the permanent magnet, and $V$ is its volume. Suppose, we're working with an industrial grade neodymium bar, IMNB1074 to be precise. Given its dimensions of 0.5 in x0.06 in x 0.13 in and remanence of 13200 Gauss, it's magnetic moment is:
$\vec{m} = {{1}\over{\mu_0}} \vec{B_r} V = (1.26\times10^{-6}) (1.48 T) (6.55\times 10^-5 m^3) = 77 Am^2 $
Earth's Magnetic Moment
Since the magnetic field is generated by the geodynamo of the outer core, it may be simply modeled as a spinning shell of charge [Princeton]. Using the princeton paper cited, the magnetic field outside the shell (R>a) as a function of the magnetic moment is:
$\vec{B_s}(r) = {{3(\vec m \cdot \vec r)\vec r - \vec m}\over{r^3}} $
Here we make a simplifying assumption by focusing entirely on the magnitude of the magnetic moment. You'll see why that's okay in a minute. First shift the $r^3$ term to the other side of the equation, and rewrite $\vec B$ as $\bar B \hat B$ (Magnitude times unit vector):
$\bar B_s r^3 \hat B_s = {3(\vec m \cdot \vec r)\vec r - \vec m}$
and calculate that left hand side, using the average of the previously cited magnet field strengths of earth.
$\bar B_s r^3 \hat B_s = {1 \over 2}(25 \times 10^{-6} T + 65 \times 10^{-6} T) (6,353,000 m)^3 \hat B= 1.15 \times 10^{16} Tm^3 \hat B$
This gives us the magnitude of Earth's magnetic moment.
$1.15 \times 10^{16} \hat B = 3(\vec m \cdot \vec r)\vec r - \vec m$
This is 150 trillion times larger than that of an industrial grade neodymium magnet.
Number of Neodymium magnets needed to simulate Earth's Magnetic Field
For completeness, I'll try to answer this when I get home tonight (I'm at work atm); but, as I'm sure you can see, artificially simulating Earth's magnetic field require an astronomical number of magnets.
