I hope I understood the question correctly. They try to compute exactly the same thing, so they really are equivalent. I'll use Chebyshev polynomials because they are easy to analyze.
Given a function $f(x)$ on $[-1,1]$, the spectral interpolant is the truncation of
$$ \begin{aligned}
f(x) &= \sum_{n\geq0} \bar a_n T_n(x), \\
\bar a_n &= \frac{1+[n>0]}{\pi}\int_{-1}^1 f(x)T_n(x)\frac{dx}{\sqrt{1-x^2}} \\&= \frac{1+[n>0]}{\pi} \int_{0}^\pi f(\cos\theta)\cos(n\theta)\,\mathrm{d}\theta.
\end{aligned} $$
The $N$-th degree Lagrange interpolant, using the roots of $T_{N+1}(x)$ is given by
$$ a_n = \frac{1+[n>0]}{N+1}\sum_{k=0}^N f(x_k) T_n(x_k), \qquad x_k = \cos\frac{\pi (k+\frac12)}{N+1}. $$
This uses the fact that $\sum_{k=0}^N T_m(x_k)T_n(x_k)$ is zero when $m\neq n$, $m,n\leq N$.
The formula for $a_n$ is nothing but the discrete cosine transform (type-II) applied to the function values at $x_k$, due to $T_n(x_k) = \cos \pi n(k+\frac12)/(N+1)$.
These are not, strictly speaking, the same, though.
The formula for $a_n$ is a trapezoidal rule approximation to the Fourier cosine integral in the formula for the exact coefficients $\bar a_n$. The trapezoidal rule is known to be exponentially accurate for smooth periodic functions (Trefethen-Weideman 2014), which $f(\cos\theta)$ is. Since for a spectral interpolant you would still have to evaluate the integral somehow, the Lagrange interpolant with the roots as nodes is just a way of evaluating that integral.
I tried to compute the exact difference between $a_n$ and $\bar a_n$, by expanding the sum for $a_n$ using the full series for $f$, and using the identity
$$ \sum_{k=0}^{N} \cos(j\theta_k)\cos(n\theta_k) = \frac{N+1}{2}\big( [N+1\setminus j-n](-1)^{(j-n)N/(N+1)} + [N+1\setminus j+n](-1)^{(j+n)N/(N+1)}\big), $$
and for a complex-differentiable function $g(\theta)=f(\cos\theta)$ that is holomorphic in the region of the complex plane $|\Im \theta|<\alpha$, heuristically the error appears to be something on the order
$$ a_n - \bar a_n \sim |\bar a_{N+1-n}| \lesssim e^{-\alpha(N+1-n)}. $$
So for smooth functions this error decays very quickly and becomes negligible, so the Lagrange and the spectral interpolants can be considered identical.
Edit. What is the relationship between $\sum a_nT_n(x)$ and $\sum f(x_k) \ell_k(x)$?
Let $a_n$ be defined as above, let $f_1(x) = \sum_{n=0}^{N} a_n T_n(x)$, and let $f_2(x) = \sum_{k=0}^N f(x_k) \ell_k(x)$, where $\ell_k(x) = \prod_{j\neq k} (x-x_j)/(x_k-x_j)$.
Both $f_1(x)$ and $f_2(x)$ are polynomials in $x$ of degree $N$, by construction.
Using the above definition of $a_n$, together with
$$T_n(x_k) = \cos(n\theta_k), \qquad \theta_k = \pi(k+\tfrac12)/(N+1) $$
we can check that $f_1(x_k) = f(x_k)$, using the identity
$$ \sum_{n=0}^{N} \frac{1+[n>0]}{N+1} \cos(n\theta_j)\cos(n\theta_k) = [j=k]. $$
Therefore $f_1(x)$ and $f_2(x)$ are (non-identically-zero) polynomials of degree $N$ that pass through the same $N+1$ points, and therefore are the same polynomial.
Using the form of the interpolating polynomial in terms of $a_n$ makes the relationship with the Chebyshev series of the function $f(x)$ clearer than the Lagrange interpolation form.