Edited to clarify question and give accepted answer in context.
Looking at the table of Fourier Transforms on Wikibooks line 14
gives the Fourier transform of the triangle function
$\left(1-\frac{2|t|}{\tau}\right)\mathrm{rect}\left(\frac{t}{\tau}\right)$
as
$\frac{\tau}{2}\mathrm{sinc}^2\left(\frac{\tau\omega}{4\pi}\right)$.
The triangle function is the convolution of two rectangle functions $\mathrm{rect}\left(\frac{t}{\tau/2}\right)$.
Each rectangle function, by Line 12 has a Fourier transform $\frac{\tau}{2}\mathrm{sinc}\left(\frac{\tau\omega}{4\pi}\right)$.
The Fourier transform of the convolution should be the squared term $\frac{\tau^2}{4}\mathrm{sinc}^2\left(\frac{\tau\omega}{4\pi}\right)$ which is off by a factor of $\tau/2$.
Answer by @dfnu
The convolution of the two rectangle functions is the triangle function multiplied by $\tau/2$; which is the area of the product of the two rectangles for zero lag. To get the triangle function given, the convolution must be divided by $\tau/2$.
Best Answer
I think you are missing factor $\frac2\tau$.
If you convolve $\operatorname{rect}\left(\frac{2t}\tau\right)$ with itself, you'll obtain a Fourier transform equal to $\frac{\tau^2}{4}\operatorname{sinc}^2\left(\frac{\tau\omega}{4\pi}\right)$.
However, this convolution does not give as a result the triangular function your mentioning. The convolution in $0$ is equal to the area of the rectangle, i.e. $\frac\tau 2$, whereas the triangle in $0$ is equal to $1$. So the triangle is actually \begin{eqnarray} h(t) &=&\left(1-\frac{2|t|}{\tau}\right)\mathrm{rect}\left(\frac{t}{\tau}\right)=\\ &=& \frac2{\tau}\cdot \operatorname{rect}\left(\frac{2t}\tau\right) *\operatorname{rect}\left(\frac{2t}\tau\right), \end{eqnarray}
and its Fourier transform is $$H(\omega) =\frac2\tau \frac{\tau^2}{4}\operatorname{sinc}^2\left(\frac{\tau\omega}{4\pi}\right).$$