Edit: As Matthias pointed out, the following argument only works for the ball with radius 1/2.

To measure volume, we need to agree on a unit of volume [1]. The traditional way of doing this is to set the volume of the unit cube to one.

Now, think about the $n$-ball inscribed in the unit $n$-cube. As we increase $n$, the ball's diameter stays constant, but what happens to its volume? When $n = 1$, the ball takes up the whole unit cube, so its volume is one. When $n = 2$, the ball no longer takes up the whole unit cube, so its volume is less than one. When $n = 3$, the ball takes up even less of the unit cube, so its volume is even smaller.

There's an easy way to see that when you go from $\mathbb{R}^n$ to $\mathbb{R}^{n + 1}$, the fraction of the unit cube occupied by the inscribed ball goes down. Start with an $n$-ball inscribed in the unit $n$-cube, and extrude both objects into the $(n + 1)$st dimenion. Now you have an $(n + 1)$-cylinder inscribed in an $(n + 1)$-cube. The fraction of the $(n + 1)$-cube occupied by the $(n + 1)$-cylinder is clearly the same as the fraction of the $n$-cube occupied by the $n$-ball. It's easy to see, however, that the $(n + 1)$-ball inscribed in the $(n + 1)$-cube fits inside the inscribed $(n + 1)$-cylinder with room to spare.

This argument only shows that the volume of the unit-diameter $n$-ball decreases as $n$ grows; it doesn't show that the volume goes to zero. I'm hopeful, however, that a more sophisticated version of the same argument might do the trick!

Edit: A more sophisticated version of the same argument does do the trick, and Matthias posted it while I was writing my post! Hooray!

[1] To be more sophisticated about it: the differential n-form in $\mathbb{R}^n$ is only unique up to multiplication by a constant, so we need to settle on a constant.

Now, think about the $n$-ball inscribed in the unit $n$-cube. As we increase $n$, the ball's diameter stays constant, but what happens to its volume? When $n = 1$, the ball takes up the whole unit cube, so its volume is one. When $n = 2$, the ball no longer takes up the whole unit cube, so its volume is less than one. When $n = 3$, the ball takes up even less of the unit cube, so its volume is even smaller.
There's an easy way to see that when you go from $\mathbb{R}^n$ to $\mathbb{R}^{n + 1}$, the fraction of the unit cube occupied by the inscribed ball goes down. Start with an $n$-ball inscribed in the unit $n$-cube, and extrude both objects into the $(n + 1)$st dimenion. Now you have an $(n + 1)$-cylinder inscribed in an $(n + 1)$-cube. The fraction of the $(n + 1)$-cube occupied by the $(n + 1)$-cylinder is clearly the same as the fraction of the $n$-cube occupied by the $n$-ball. It's easy to see, however, that the $(n + 1)$-ball inscribed in the $(n + 1)$-cube fits inside the inscribed $(n + 1)$-cylinder with room to spare.
This argument only shows that the volume of the unit-diameter $n$-ball decreases as $n$ grows; it doesn't show that the volume goes to zero. I'm hopeful, however, that a more sophisticated version of the same argument might do the trick!
[1] To be more sophisticated about it: the differential n-form in $\mathbb{R}^n$ is only unique up to multiplication by a constant, so we need to settle on a constant.