Here is a (perhaps) more elementary method. Let $X$ be the amount of numbers you need to add until the sum exceeds $1$. Then (by linearity of expectation):
$$ \mathbb{E}[X] = 1 + \sum_{k \geq 1} \Pr[X > k] $$
Now $X > k$ if the sum of the first $k$ numbers $x_1,\ldots,x_k$ is smaller than $1$. This is exactly equal to the volume of the $k$-dimensional set:
$$ \left\{(x_1,\ldots,x_k) : \sum_{i=1}^k x_i \leq 1, \, x_1,\ldots,x_k \geq 0\right\}$$
This is known as the $k$-dimensional simplex. When $k = 1$, we get a line segment of length $1$. When $k = 2$, we get a right isosceles triangle with sides of length $1$, so the area is $1/2$. When $k=3$, we get a triangular pyramid (tetrahedron) with unit sides, so the volume is $1/6$. In general, the volume is $1/k!$, and so
$$ \mathbb{E}[X] = 1 + \sum_{k \geq 1} \frac{1}{k!} = e. $$
Let $u(n)$ be the expected value of the first roll that makes the total $\ge n$.
Thus $u(n) = 7/2$ for $n \le 1$. But for $2 \le n \le 6$, conditioning on the first roll
we have $$u(n) = \left( \sum_{j=1}^{n-1} u(n-j)
+ \sum_{j=n}^{6} j \right)/6$$
That makes
$$ u_{{2}}={\frac {47}{12}},u_{{3}}={\frac {305}{72}},u_{{4}}={\frac {1919}{432}},u_{{5}}={\frac {11705}{2592}},u_{{6}}={\frac {68975}{15552
}}$$
And then for $n > 6$, again conditioning on the first roll,
$$u(n) = \frac{1}{6} \sum_{j=1}^6 u(n-j)$$
The result is
$$u(51) = \frac {7005104219281602775658473799867927981609}{1616562554929528121286279200913072586752} \approx 4.333333219$$
It turns out that as $n \to \infty$, $u(n) \to 13/3$.
EDIT: Note that the
general solution to the recurrence
$\displaystyle u(n) = \frac{1}{6} \sum_{j=1}^6 u(n-j)$
is
$$ u(n) = c_0 + \sum_{j=1}^5 c_j r_j^n$$
where $r_j$ are the roots of
$$\dfrac{6 r^6 - (1 + r + \ldots + r^5)}{r - 1} = 6 r^5 + 5 r^4 + 4 r^3 + 3 r^2 + 2 r + 1 = 0$$
Those all have absolute value $< 1$, so $\lim_{n \to \infty} u(n) = c_0$. Now
$6 u(n+5) + 5 u(n+4) + \ldots + u(n) = (6 + 5 + \ldots + 1) c_0 = 21 c_0$
because the terms in each $r_j$ vanish. Taking $n = 1$ with the values of
$u_1$ to $u_6$ above gives us
$c_0 = 13/3$.
Best Answer
Your intuition is formalised by Wald's equation. Apply it with $X_n$ equal to the $n$-th random number and $N=\inf\left\{n\in\mathbb N^*\mid\sum_{k=1}^nX_k\ge1\right\}$. Then $$ \mathbb E\left[\sum_{k=1}^NX_k\right]=\mathbb E[N]\times\mathbb E[X_1]=\mathrm{e}\times\frac12. $$