This answer formalizes Rahul's clever answer, and shows how it works because they have currently gotten away with (wonderfully slick and probably accidental) murder. In particular, their argument about the constraints doesn't make any sense because we could have $x = 1/3, y=1/2, z=1$ such that the length would be $3$, not $1$. And why should we be able to normalize by the expected value of the length over this region?
The initial setup was clever. Effectively, we want to know what the expected value of $M = x+y+z$ is, subject to the constraint that $3x + 2y + z = 1$ and all non-negative, which is a planar triangular region in $\mathbb{R}_+^3$ (i.e. where $x,y,z\geq 0$). But to make the math even nicer, we can compute the expected value of $M$ on the planar region given by $3x + 2y + z = L$ for any $L>0$, and then just normalize by $L$. This is allowed by Rahul's renormalization explanation, or by the linearity of $M$.
By the linearity of expectation, $\mathbb{E}M = \mathbb{E}X + \mathbb{E}Y + \mathbb{E}Z$, where $X,Y,Z$ are the coordinate functions in this region. So how do we go about computing these expectations in this region? Do we need to integrate?
Well, no. This region is a triangle, so we can just calculate the centroid by averaging the vertices. Notice that the vertices are
$$
\mathbf{v_1} = \begin{pmatrix}
L/3 \\ 0 \\ 0
\end{pmatrix}, \mathbf v_2 = \begin{pmatrix}
0 \\ L/2 \\ 0
\end{pmatrix}, \mathbf v_3 = \begin{pmatrix}
0 \\ 0 \\ L
\end{pmatrix},
$$
so the centroid is
$$
\frac{\mathbf v_1 + \mathbf v_2 + \mathbf v_3}{3} = \frac{L}{18}\begin{pmatrix}
2 \\ 3 \\ 6
\end{pmatrix}.
$$
Of course now we are done, because we plug in $L =1$ and evaluate $M$ to get $\frac{11}{18}$, and if we want the smallest it's just $\mathbb{E}X = \frac{2}{18} = \frac{1}{9}$.
But let's figure out why Rahul's trick worked. Observe that Rahul basically found the center of the box $[0,2n]\times[0,3n]\times[0,6n]$. They found these bounds the same way that we found the vertices of our triangle, but made the mistake of treating the box as the proper region, not the triangle. Then they got their coordinates, $(n, 1.5n, 3n)^T$, and said well, these coordinates correspond to a length of $9n$, so normalizing by this they got the result. However, notice that the vertices with at $2n, 3n, 6n$ correspond to a length of $18n$, so something fishy is going on here.
What's happened is this: the centroid of the triangle with vertices $$
\mathbf{v_1} = \begin{pmatrix}
a \\ 0 \\ 0
\end{pmatrix}, \mathbf v_2 = \begin{pmatrix}
0 \\ b \\ 0
\end{pmatrix}, \mathbf v_3 = \begin{pmatrix}
0 \\ 0 \\ c
\end{pmatrix},
$$
is
$$
\frac{\mathbf v_1 + \mathbf v_2 + \mathbf v_3}{3} = \frac{1}{3}\begin{pmatrix}
a \\ b \\ c
\end{pmatrix},
$$
as we discussed above. However, the center of the box $[0,a]\times[0,b]\times[0,c]$ is
$$
\frac{\mathbf v_1 + \mathbf v_2 + \mathbf v_3}{2} = \frac{1}{2}\begin{pmatrix}
a \\ b \\ c
\end{pmatrix}.
$$
This is the centroid given by scaling the triangle by $3/2$. For our lengths problems, this means Rahul found the centroid of the triangle with $3/2$ the length of the one implied by their setup. But because they renormalized, this didn't end up mattering. However, we have to admit they got kind of lucky here.
Anyways, the cool thing about this method is that it easily generalized to $n$ cuts. Now our pieces have length $x_1, x_1 + x_2, \dots, x_1 + \dots + x_n$. This means the total length is $nx_1 + (n-1)x_2 + \dots + x_1$, so our space is the $n$-simplex in the $n-1$ dimensional hyperplane in $\mathbb{R}^n_+$ given by
$$
nx_1 + (n-1)x_2 + \dots + x_1 = 1,
$$
which has vertices
$$
\begin{pmatrix}
\frac{1}{n} \\ 0 \\ \vdots \\ 0
\end{pmatrix},
\begin{pmatrix}
0 \\ \frac{1}{n-1} \\ \vdots \\ 0
\end{pmatrix}, \dots, \begin{pmatrix}
0 \\ 0 \\ \vdots \\ 1
\end{pmatrix}.
$$
Thus the centroid is
$$
\frac{1}{n}\begin{pmatrix}
\frac{1}{n} \\ \frac{1}{n-1} \\ \vdots \\ 1
\end{pmatrix},
$$
so the expected largest piece is
$$
\frac{1}{n}\left(\frac{1}{n} + \frac{1}{n-1} + \dots + 1 \right) = \frac{H_n}{n},
$$
as per Mike's answer. The expected smallest piece has size $\frac{1}{n^2}$.
