Zero, First, Second ... nth-order Hold
This is not the case. First of all, a second-order hold would use three sample points to compute an interpolation polynomial, but your suggested impulse response $\text{tri}(t)\star\text{tri}(t)$ is non-zero in an interval of size $4$ (assuming a sample interval of $T=1$, as you do in your question). However, the impulse response corresponding to a second-order hold must have a support of length $3$.

Now you could suggest that an $n^{th}$-order hold could have an impulse response which is the convolution of $n$ rectangular functions. In this case you would get the correct support size, but of course that's not sufficient.

An $n^{th}$-order hold computes a piece-wise interpolation using $n+1$ consecutive data points. This is in analogy with a zero-order hold using a single data point, and a first-order hold, which uses two data points. This definition is commonly used in the literature (see e.g. here and here).

It's straightforward to show that the second-order polynomial that interpolates three data points $y[-1]$, $y[0]$, and $y[1]$ is given by

$$P(t)=y[-1]\frac{t(t-1)}{2}+y[0](1-t^2)+y[1]\frac{t(t+1)}{2}\tag{1}$$

In order to find the impulse response achieving the interpolation given by $(1)$, we have to equate $(1)$ with the expression

$$y[-1]h(t+1)+y[0]h(t)+y[1]h(t-1)\tag{2}$$

If we choose the support of the impulse response $h(t)$ as the interval $[-1,2]$, which is equivalent to choosing the interpolation interval $[0,1]$, equating $(1)$ and $(2)$ results in the following impulse response of a second-order hold:

$$h(t)=\begin{cases}\frac12(t+1)(t+2),&amp;\quad -1&lt;t&lt;0\\
1-t^2,&amp;\quad0\le t \le 1\\
\frac12 (t-1)(t-2),&amp;\quad 1&lt;t&lt;2\\
0,&amp;\quad\text{otherwise}\end{cases}\tag{3}$$

The impulse response $(3)$ of a second-order hold looks like this:


I leave it up to you to show that this impulse response cannot be generated by convolving three rectangular functions with each other.