Obviously, N must be
seven, because 7 is the only number that gives a product ending in 9 when multiplied by 7.
Then, let's take a guess to get acquainted with the types of trouble we are going to face. We get the simplest calculations if we set E=0, so let's try that.
Looks like we'll then need three multiplications (x * 7 = yz) with all the digits unique. That seems easy:
$$ 7 \times 7 = 49 $$
$$ 8 \times 7 = 56 $$
$$ 3 \times 7 = 21 $$
It seems we have accidentally stumbled on a solution. Even more than that, since we have zeroes in all the strategically important spots, the order in which we place the multiplications doesn't matter. This means there are at least two solutions:
$$ 80307 \times 7 = 562149 $$
$$ 30807 \times 7 = 215649 $$
A double solution doesn't really seem Gardner's style though. I'll try and see if I can't find the original..
EDIT: Found it. The original has the extra stipulation that "SEVEN" must be divisible by 7, which rules out the bigger answer. Looks like SciAm forgot to include that extra condition in their 2014 reprint of the puzzle.
Source: "Index to Mathematical Problems 1975-1979, edited by Stanley Rabinowitz, Mark Bowron", highlighting mine.